CS计算机代考程序代写 matlab algorithm Linear Algebra Module II: Vector Spaces and Linear Transformations

Linear Algebra Module II: Vector Spaces and Linear Transformations

Linear Algebra Module II: Vector Spaces and Linear
Transformations

Summer 2021

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 1 / 73

Table of Contents

1 Vector Spaces
The space Rn
Vector Spaces – Definition
Subspaces
Spanning Sets, Row Spaces, Column Spaces
Nullspaces
Range
Bases and Dimension
Coordinate Systems

2 Linear Transformations
Linear Transformations – Definition, properties, and terminology
Linear Transformations – Matrix representation
Change of basis

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 2 / 73

Table of Contents

1 Vector Spaces
The space Rn
Vector Spaces – Definition
Subspaces
Spanning Sets, Row Spaces, Column Spaces
Nullspaces
Range
Bases and Dimension
Coordinate Systems

2 Linear Transformations
Linear Transformations – Definition, properties, and terminology
Linear Transformations – Matrix representation
Change of basis

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 3 / 73

Vector Spaces – Overview

A vector space is a set equipped with two operations, vector addition and
scalar multiplication, satisfying certain properties extending those for the
real (or complex) numbers. Vector spaces, and their subspaces, are the
basic objects of Linear Algebra.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 4 / 73

The space Rn

Recall that Rm×n denotes the set of all m × n matrices with real entries,
and that the elements of this set are called row resp. column vectors when
m = 1 resp. n = 1.

Definition of Rn

Rn is the space Rn×1. In other words, a real n-dimensional vector will
always refer to an n × 1 real column vector.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 5 / 73

The space Rn

Recall that Rm×n denotes the set of all m × n matrices with real entries,
and that the elements of this set are called row resp. column vectors when
m = 1 resp. n = 1.

Definition of Rn

Rn is the space Rn×1. In other words, a real n-dimensional vector will
always refer to an n × 1 real column vector.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 5 / 73

The space Rn

From the definition of matrix addition and scalar multiplication, we see
that in Rn we can

i) add two vectors together, and

ii) multiply a vector by a scalar

with the result being a (possibly different) vector in the same space that
we started with. In other words

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 6 / 73

The space Rn

From the definition of matrix addition and scalar multiplication, we see
that in Rn we can

i) add two vectors together, and

ii) multiply a vector by a scalar

with the result being a (possibly different) vector in the same space that
we started with. In other words

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 6 / 73

The space Rn

From the definition of matrix addition and scalar multiplication, we see
that in Rn we can

i) add two vectors together, and

ii) multiply a vector by a scalar

with the result being a (possibly different) vector in the same space that
we started with. In other words

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 6 / 73

The space Rn

From the definition of matrix addition and scalar multiplication, we see
that in Rn we can

i) add two vectors together, and

ii) multiply a vector by a scalar

with the result being a (possibly different) vector in the same space that
we started with. In other words

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 6 / 73

The space Rn

Closure

[C1] (Closure under vector addition) Given v,w ∈ Rn, v + w ∈ Rn.
[C2] (Closure under scalar multiplication) Given v ∈ Rn and α ∈ R,
αv ∈ Rn.

Moreover, the space Rn equipped with these two operations satisfies
certain fundamental properties. In what follows, u, v, w denote arbitrary
vectors in Rn, while α, β represent arbitrary scalars in R.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 7 / 73

The space Rn

Closure
[C1] (Closure under vector addition) Given v,w ∈ Rn, v + w ∈ Rn.

[C2] (Closure under scalar multiplication) Given v ∈ Rn and α ∈ R,
αv ∈ Rn.

Moreover, the space Rn equipped with these two operations satisfies
certain fundamental properties. In what follows, u, v, w denote arbitrary
vectors in Rn, while α, β represent arbitrary scalars in R.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 7 / 73

The space Rn

Closure
[C1] (Closure under vector addition) Given v,w ∈ Rn, v + w ∈ Rn.
[C2] (Closure under scalar multiplication) Given v ∈ Rn and α ∈ R,
αv ∈ Rn.

Moreover, the space Rn equipped with these two operations satisfies
certain fundamental properties. In what follows, u, v, w denote arbitrary
vectors in Rn, while α, β represent arbitrary scalars in R.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 7 / 73

The space Rn

Closure
[C1] (Closure under vector addition) Given v,w ∈ Rn, v + w ∈ Rn.
[C2] (Closure under scalar multiplication) Given v ∈ Rn and α ∈ R,
αv ∈ Rn.

Moreover, the space Rn equipped with these two operations satisfies
certain fundamental properties. In what follows, u, v, w denote arbitrary
vectors in Rn, while α, β represent arbitrary scalars in R.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 7 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.

[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).

[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.

[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.

[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.

[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.

[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).

[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

The 8 basic properties of Rn

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 8 / 73

The space Rn

Why does Rn satisfies these properties?

First, the definition of matrix addition and scalar multiplication imply [C1]
and [C2].

Next, the properties [A1] – [A8], excepting [A3] and [A4], are a
consequence of the Main Theorem of Matrix Algebra . The so-called
existential properties [A3] and [A4] (referring to the fact they claim the
existence of certain vectors) follow by direct observation.

These properties isolate the fundamental algebraic structure of Rn, and
lead to the following definition of a vector space over R (one of the most
central in all of linear algebra).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 9 / 73

The space Rn

Why does Rn satisfies these properties?

First, the definition of matrix addition and scalar multiplication imply [C1]
and [C2].

Next, the properties [A1] – [A8], excepting [A3] and [A4], are a
consequence of the Main Theorem of Matrix Algebra . The so-called
existential properties [A3] and [A4] (referring to the fact they claim the
existence of certain vectors) follow by direct observation.

These properties isolate the fundamental algebraic structure of Rn, and
lead to the following definition of a vector space over R (one of the most
central in all of linear algebra).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 9 / 73

The space Rn

Why does Rn satisfies these properties?

First, the definition of matrix addition and scalar multiplication imply [C1]
and [C2].

Next, the properties [A1] – [A8], excepting [A3] and [A4], are a
consequence of the Main Theorem of Matrix Algebra . The so-called
existential properties [A3] and [A4] (referring to the fact they claim the
existence of certain vectors) follow by direct observation.

These properties isolate the fundamental algebraic structure of Rn, and
lead to the following definition of a vector space over R (one of the most
central in all of linear algebra).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 9 / 73

The space Rn

Why does Rn satisfies these properties?

First, the definition of matrix addition and scalar multiplication imply [C1]
and [C2].

Next, the properties [A1] – [A8], excepting [A3] and [A4], are a
consequence of the Main Theorem of Matrix Algebra . The so-called
existential properties [A3] and [A4] (referring to the fact they claim the
existence of certain vectors) follow by direct observation.

These properties isolate the fundamental algebraic structure of Rn, and
lead to the following definition of a vector space over R (one of the most
central in all of linear algebra).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 9 / 73

Vector Spaces – Definition

A real vector space is a non-empty set V equipped with two operations –
vector addition “+” and scalar multiplication “·”- which satisfy C1, C2 as
well as A1 – A8:

Closure Axioms

[C1] (Closure under vector addition) Given v,w ∈ V , v + w ∈ V .
[C2] (Closure under scalar multiplication) Given v ∈ V and α ∈ R,
αv ∈ V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 10 / 73

Vector Spaces – Definition

A real vector space is a non-empty set V equipped with two operations –
vector addition “+” and scalar multiplication “·”- which satisfy C1, C2 as
well as A1 – A8:

Closure Axioms
[C1] (Closure under vector addition) Given v,w ∈ V , v + w ∈ V .

[C2] (Closure under scalar multiplication) Given v ∈ V and α ∈ R,
αv ∈ V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 10 / 73

Vector Spaces – Definition

A real vector space is a non-empty set V equipped with two operations –
vector addition “+” and scalar multiplication “·”- which satisfy C1, C2 as
well as A1 – A8:

Closure Axioms
[C1] (Closure under vector addition) Given v,w ∈ V , v + w ∈ V .
[C2] (Closure under scalar multiplication) Given v ∈ V and α ∈ R,
αv ∈ V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 10 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms

[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.

[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).

[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.

[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.

[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.

[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.

[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).

[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Definition
For u, v, w arbitrary vectors in V , and α, β arbitrary scalars in R,

The Vector Space Axioms
[A1] (Commutativity of addition) v + w = w + v.
[A2] (Associativity of addition) (u + v) + w = u + (v + w).
[A3] (Existence of a zero vector) There is a vector z with
z + v = v + z = v.
[A4] (Existence of additive inverses) For each v, there is a vector −v
with v + (−v) = (−v) + v = z.
[A5] (Distributivity of scalar multiplication over vector addition)
α(v + w) = αv + αw.
[A6] (Distributivity of scalar addition over scalar multiplication)
(α + β)v = αv + βv.
[A7] (Associativity of scalar multiplication) (αβ)v = α(βv).
[A8] (Scalar multiplication with 1 is the identity) 1v = v.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 11 / 73

Vector Spaces – Examples

A vector space can properly be represented as a triple (V ,+, ·), to
emphasize the fact that the algebraic structure depends not just on the
underlying set of vectors, but on the choice of operations representing
addition and scalar multiplication.

Examples

Let V = Rm×n, the space of m × n matrices, with addition given by
matrix addition and scalar multiplication as previously defined for matrices.

Then (Rm×n,+, ·) is a vector space. Again, as with Rn, the closure axioms
are seen to be satisfied as a direct consequence of the definitions, while
the other properties follow from the Main Theorem of Matrix Algebra,
together with direct construction of the m × n “zero vector” 0m×n, as well
as additive inverses as indicated in [A4].

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 12 / 73

Vector Spaces – Examples

A vector space can properly be represented as a triple (V ,+, ·), to
emphasize the fact that the algebraic structure depends not just on the
underlying set of vectors, but on the choice of operations representing
addition and scalar multiplication.

Examples

Let V = Rm×n, the space of m × n matrices, with addition given by
matrix addition and scalar multiplication as previously defined for matrices.

Then (Rm×n,+, ·) is a vector space. Again, as with Rn, the closure axioms
are seen to be satisfied as a direct consequence of the definitions, while
the other properties follow from the Main Theorem of Matrix Algebra,
together with direct construction of the m × n “zero vector” 0m×n, as well
as additive inverses as indicated in [A4].

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 12 / 73

Vector Spaces – Examples

A vector space can properly be represented as a triple (V ,+, ·), to
emphasize the fact that the algebraic structure depends not just on the
underlying set of vectors, but on the choice of operations representing
addition and scalar multiplication.

Examples

Let V = Rm×n, the space of m × n matrices, with addition given by
matrix addition and scalar multiplication as previously defined for matrices.
Then (Rm×n,+, ·) is a vector space.

Again, as with Rn, the closure axioms
are seen to be satisfied as a direct consequence of the definitions, while
the other properties follow from the Main Theorem of Matrix Algebra,
together with direct construction of the m × n “zero vector” 0m×n, as well
as additive inverses as indicated in [A4].

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 12 / 73

Vector Spaces – Examples

A vector space can properly be represented as a triple (V ,+, ·), to
emphasize the fact that the algebraic structure depends not just on the
underlying set of vectors, but on the choice of operations representing
addition and scalar multiplication.

Examples

Let V = Rm×n, the space of m × n matrices, with addition given by
matrix addition and scalar multiplication as previously defined for matrices.
Then (Rm×n,+, ·) is a vector space. Again, as with Rn, the closure axioms
are seen to be satisfied as a direct consequence of the definitions, while
the other properties follow from the Main Theorem of Matrix Algebra,
together with direct construction of the m × n “zero vector” 0m×n, as well
as additive inverses as indicated in [A4].

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 12 / 73

Vector Spaces – Examples

Examples

F [a, b] = the space of real-valued functions on the closed interval [a, b]

The sum of two functions is defined by the equality

(f + g)(x) := f (x) + g(x)

while the scalar multiple αf of the function f is defined by

(αf )(x) := α(f (x))

Theorem
Equipped with addition and scalar multiplication as just defined,
(F [a, b],+, ·) is a vector space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 13 / 73

Vector Spaces – Examples

Examples

F [a, b] = the space of real-valued functions on the closed interval [a, b]

The sum of two functions is defined by the equality

(f + g)(x) := f (x) + g(x)

while the scalar multiple αf of the function f is defined by

(αf )(x) := α(f (x))

Theorem
Equipped with addition and scalar multiplication as just defined,
(F [a, b],+, ·) is a vector space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 13 / 73

Vector Spaces – Examples

Examples

F [a, b] = the space of real-valued functions on the closed interval [a, b]

The sum of two functions is defined by the equality

(f + g)(x) := f (x) + g(x)

while the scalar multiple αf of the function f is defined by

(αf )(x) := α(f (x))

Theorem
Equipped with addition and scalar multiplication as just defined,
(F [a, b],+, ·) is a vector space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 13 / 73

Vector Spaces – Examples

Examples

F [a, b] = the space of real-valued functions on the closed interval [a, b]

The sum of two functions is defined by the equality

(f + g)(x) := f (x) + g(x)

while the scalar multiple αf of the function f is defined by

(αf )(x) := α(f (x))

Theorem
Equipped with addition and scalar multiplication as just defined,
(F [a, b],+, ·) is a vector space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 13 / 73

Subspaces – Definition

If V = {V ,+, ·} is a vector space, then W ⊆ V is called a subspace of V
if it has at least one element (is non-empty) and the restriction to W of
the sum and scalar product operations of V make W a vector space.

The following theorem tells us exactly when a subset of V is a subspace.

Theorem
Let W be a non-empty subset of vectors in V , which satisfy the two
closure axioms C1 and C2 with respect to the operations on V . Then W
is a subspace of V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 14 / 73

Subspaces – Definition

If V = {V ,+, ·} is a vector space, then W ⊆ V is called a subspace of V
if it has at least one element (is non-empty) and the restriction to W of
the sum and scalar product operations of V make W a vector space.

The following theorem tells us exactly when a subset of V is a subspace.

Theorem
Let W be a non-empty subset of vectors in V , which satisfy the two
closure axioms C1 and C2 with respect to the operations on V . Then W
is a subspace of V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 14 / 73

Subspaces – Definition

If V = {V ,+, ·} is a vector space, then W ⊆ V is called a subspace of V
if it has at least one element (is non-empty) and the restriction to W of
the sum and scalar product operations of V make W a vector space.

The following theorem tells us exactly when a subset of V is a subspace.

Theorem
Let W be a non-empty subset of vectors in V , which satisfy the two
closure axioms C1 and C2 with respect to the operations on V . Then W
is a subspace of V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 14 / 73

Subspaces

Examples
For a closed interval [a, b] ⊂ R, let C [a, b] ⊂ F [a, b] denote the subset of
continuous (real-valued) functions on [a, b]. Then C [a, b] is a subspace of
F [a, b].

Reason: First, C [a, b] contains the zero function on [a, b], so it is
non-empty. And C [a, b] is closed under sum and scalar products (these
results were verified in first-term Calculus). So by the above theorem
C [a, b] is a subspace of F [a, b].

Examples
Let P∞ be the set of polynomials in x with real coefficients, and let
Pn ⊂ P∞ be the subset of P∞ consisting of polynomials in x of degree less
than n (n ≥ 1). Show that P∞ is a subspace of F (−∞,∞), and that Pn
is a subspace of P∞ for all n ≥ 1.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 15 / 73

Subspaces

Examples
For a closed interval [a, b] ⊂ R, let C [a, b] ⊂ F [a, b] denote the subset of
continuous (real-valued) functions on [a, b]. Then C [a, b] is a subspace of
F [a, b].

Reason: First, C [a, b] contains the zero function on [a, b], so it is
non-empty. And C [a, b] is closed under sum and scalar products (these
results were verified in first-term Calculus). So by the above theorem
C [a, b] is a subspace of F [a, b].

Examples
Let P∞ be the set of polynomials in x with real coefficients, and let
Pn ⊂ P∞ be the subset of P∞ consisting of polynomials in x of degree less
than n (n ≥ 1). Show that P∞ is a subspace of F (−∞,∞), and that Pn
is a subspace of P∞ for all n ≥ 1.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 15 / 73

Subspaces

Examples
For a closed interval [a, b] ⊂ R, let C [a, b] ⊂ F [a, b] denote the subset of
continuous (real-valued) functions on [a, b]. Then C [a, b] is a subspace of
F [a, b].

Reason: First, C [a, b] contains the zero function on [a, b], so it is
non-empty. And C [a, b] is closed under sum and scalar products (these
results were verified in first-term Calculus). So by the above theorem
C [a, b] is a subspace of F [a, b].

Examples
Let P∞ be the set of polynomials in x with real coefficients, and let
Pn ⊂ P∞ be the subset of P∞ consisting of polynomials in x of degree less
than n (n ≥ 1). Show that P∞ is a subspace of F (−∞,∞), and that Pn
is a subspace of P∞ for all n ≥ 1.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 15 / 73

Linear Combinations – Definition

The operation of forming linear combinations of vectors is at the heart of
Linear Algebra; it is, arguably, the central construct of the entire subject.
And yet, it is relatively straightforward to describe.

Definition
Let (V ,+, ·) be a vector space, and v1, . . . , vn ∈ V a collection of n
vectors in V .

A linear combination of v1, . . . , vn is a sum of scalar
multiples of these vectors; in other words, a sum of the form

α1v1 + α2v2 + · · ·+ αnvn

for some choice of scalars α1, α2, . . . , αn.

A vector v is a linear combination of v1, . . . , vn if it can be written in this
form.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 16 / 73

Linear Combinations – Definition

The operation of forming linear combinations of vectors is at the heart of
Linear Algebra; it is, arguably, the central construct of the entire subject.
And yet, it is relatively straightforward to describe.

Definition
Let (V ,+, ·) be a vector space, and v1, . . . , vn ∈ V a collection of n
vectors in V . A linear combination of v1, . . . , vn is a sum of scalar
multiples of these vectors; in other words, a sum of the form

α1v1 + α2v2 + · · ·+ αnvn

for some choice of scalars α1, α2, . . . , αn.

A vector v is a linear combination of v1, . . . , vn if it can be written in this
form.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 16 / 73

Linear Combinations – Definition

The operation of forming linear combinations of vectors is at the heart of
Linear Algebra; it is, arguably, the central construct of the entire subject.
And yet, it is relatively straightforward to describe.

Definition
Let (V ,+, ·) be a vector space, and v1, . . . , vn ∈ V a collection of n
vectors in V . A linear combination of v1, . . . , vn is a sum of scalar
multiples of these vectors; in other words, a sum of the form

α1v1 + α2v2 + · · ·+ αnvn

for some choice of scalars α1, α2, . . . , αn.

A vector v is a linear combination of v1, . . . , vn if it can be written in this
form.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 16 / 73

Linear Combinations – Definition

The operation of forming linear combinations of vectors is at the heart of
Linear Algebra; it is, arguably, the central construct of the entire subject.
And yet, it is relatively straightforward to describe.

Definition
Let (V ,+, ·) be a vector space, and v1, . . . , vn ∈ V a collection of n
vectors in V . A linear combination of v1, . . . , vn is a sum of scalar
multiples of these vectors; in other words, a sum of the form

α1v1 + α2v2 + · · ·+ αnvn

for some choice of scalars α1, α2, . . . , αn.

A vector v is a linear combination of v1, . . . , vn if it can be written in this
form.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 16 / 73

Linear Combinations – Examples

Examples
Suppose v1 = [1 0 0]T , v2 = [0 1 0]T ∈ R3 (we have written these column
vectors as transposed row vectors). Then v = [2 5 0]T = 2v1 + 5v2, so v
is a linear combination of v1, v2.

On the other hand, w = [0 0 1]T is not a linear combination of v1, v2,
since the (3, 1)-entry of w is non-zero, while any linear combination of
v1, v2 would have a (3, 1)-entry equal to 0, regardless of the choice of
scalars for coefficients.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 17 / 73

Linear Combinations – Examples

Examples
Suppose v1 = [1 0 0]T , v2 = [0 1 0]T ∈ R3 (we have written these column
vectors as transposed row vectors). Then v = [2 5 0]T = 2v1 + 5v2, so v
is a linear combination of v1, v2.

On the other hand, w = [0 0 1]T is not a linear combination of v1, v2,
since the (3, 1)-entry of w is non-zero, while any linear combination of
v1, v2 would have a (3, 1)-entry equal to 0, regardless of the choice of
scalars for coefficients.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 17 / 73

Linear Independence

Linear Independence – Definition
A collection of vectors v1, . . . , vn is linearly independent if

α1v1 + α2v2 + · · ·+ αnvn = z ⇐⇒ α1 = α2 = · · · = αn = 0

In other words, the only linear combination of the vectors that produces
the zero vector is the trivial combination, where each coefficient αi = 0.

A collection of vectors v1, . . . , vn is linearly dependent iff it is not linearly
independent. More precisely

Theorem
A set of vectors v1, . . . , vn is linearly dependent iff there is a set of scalars
α1, . . . αn which are not all zero with

α1v1 + α2v2 + · · ·+ αnvn = z

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 18 / 73

Linear Independence

Linear Independence – Definition
A collection of vectors v1, . . . , vn is linearly independent if

α1v1 + α2v2 + · · ·+ αnvn = z ⇐⇒ α1 = α2 = · · · = αn = 0

In other words, the only linear combination of the vectors that produces
the zero vector is the trivial combination, where each coefficient αi = 0.

A collection of vectors v1, . . . , vn is linearly dependent iff it is not linearly
independent. More precisely

Theorem
A set of vectors v1, . . . , vn is linearly dependent iff there is a set of scalars
α1, . . . αn which are not all zero with

α1v1 + α2v2 + · · ·+ αnvn = z

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 18 / 73

Linear Independence

Linear Independence – Definition
A collection of vectors v1, . . . , vn is linearly independent if

α1v1 + α2v2 + · · ·+ αnvn = z ⇐⇒ α1 = α2 = · · · = αn = 0

In other words, the only linear combination of the vectors that produces
the zero vector is the trivial combination, where each coefficient αi = 0.

A collection of vectors v1, . . . , vn is linearly dependent iff it is not linearly
independent. More precisely

Theorem
A set of vectors v1, . . . , vn is linearly dependent iff there is a set of scalars
α1, . . . αn which are not all zero with

α1v1 + α2v2 + · · ·+ αnvn = z

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 18 / 73

Linear vs Vectorwise Independence

Call the set {v1, . . . , vn} vectorwise independent if no vector in the set
can be written as a linear combination of the remaining vectors in the set.

Theorem
A collection of vectors {v1, . . . , vn} is linearly independent iff it is
vectorwise independent.

Given a collection of vectors S = {v1, . . . vn}, a fundamental question one
can ask is whether the collection (or set) S is linearly independent. One of
our main goals in the following sections will be to develop numerical
methods for answering this question.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 19 / 73

Linear vs Vectorwise Independence

Call the set {v1, . . . , vn} vectorwise independent if no vector in the set
can be written as a linear combination of the remaining vectors in the set.

Theorem
A collection of vectors {v1, . . . , vn} is linearly independent iff it is
vectorwise independent.

Given a collection of vectors S = {v1, . . . vn}, a fundamental question one
can ask is whether the collection (or set) S is linearly independent. One of
our main goals in the following sections will be to develop numerical
methods for answering this question.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 19 / 73

Linear vs Vectorwise Independence

Call the set {v1, . . . , vn} vectorwise independent if no vector in the set
can be written as a linear combination of the remaining vectors in the set.

Theorem
A collection of vectors {v1, . . . , vn} is linearly independent iff it is
vectorwise independent.

Given a collection of vectors S = {v1, . . . vn}, a fundamental question one
can ask is whether the collection (or set) S is linearly independent. One of
our main goals in the following sections will be to develop numerical
methods for answering this question.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 19 / 73

Spanning sets, row spaces, and column spaces

The span of a set of vectors is what they generate under taking all possible
linear combinations.

Span of a finite set of vectors
If S = {v1, . . . , vn} ⊂ V is a finite collection of vectors in a vector space
V , then the span of S is the set of all linear combinations of the vectors in
S. That is

Span(S) := {α1v1 + α2v2 + · · ·+ αnvn | αi ∈ R}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 20 / 73

Spanning sets, row spaces, and column spaces

The span of a set of vectors is what they generate under taking all possible
linear combinations.

Span of a finite set of vectors
If S = {v1, . . . , vn} ⊂ V is a finite collection of vectors in a vector space
V , then the span of S is the set of all linear combinations of the vectors in
S. That is

Span(S) := {α1v1 + α2v2 + · · ·+ αnvn | αi ∈ R}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 20 / 73

Spanning sets, row spaces, and column spaces

The span of a set of vectors is what they generate under taking all possible
linear combinations.

Span of a finite set of vectors
If S = {v1, . . . , vn} ⊂ V is a finite collection of vectors in a vector space
V , then the span of S is the set of all linear combinations of the vectors in
S. That is

Span(S) := {α1v1 + α2v2 + · · ·+ αnvn | αi ∈ R}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 20 / 73

Spanning sets, row spaces, and column spaces

It will be useful to be able to discuss spans of arbitrarily large sets of
vectors.

Span of an set of vectors
If S ⊆ V is an arbitrary set of vectors in V , then

Span(S) :=

T⊆S
T finite

Span(T )

Spanning set for a vector space
If V is a vector space, and S a set of vectors in V , then we say that S is a
spanning set for V if V = Span(S).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 21 / 73

Spanning sets, row spaces, and column spaces

It will be useful to be able to discuss spans of arbitrarily large sets of
vectors.

Span of an set of vectors
If S ⊆ V is an arbitrary set of vectors in V , then

Span(S) :=

T⊆S
T finite

Span(T )

Spanning set for a vector space
If V is a vector space, and S a set of vectors in V , then we say that S is a
spanning set for V if V = Span(S).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 21 / 73

Spanning sets, row spaces, and column spaces

It will be useful to be able to discuss spans of arbitrarily large sets of
vectors.

Span of an set of vectors
If S ⊆ V is an arbitrary set of vectors in V , then

Span(S) :=

T⊆S
T finite

Span(T )

Spanning set for a vector space
If V is a vector space, and S a set of vectors in V , then we say that S is a
spanning set for V if V = Span(S).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 21 / 73

Spanning sets, row spaces, and column spaces

It will be useful to be able to discuss spans of arbitrarily large sets of
vectors.

Span of an set of vectors
If S ⊆ V is an arbitrary set of vectors in V , then

Span(S) :=

T⊆S
T finite

Span(T )

Spanning set for a vector space
If V is a vector space, and S a set of vectors in V , then we say that S is a
spanning set for V if V = Span(S).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 21 / 73

Spanning sets, row spaces, and column spaces

Theorem
If S is a non-empty subset of vectors in a vector space V , then Span(S) is
a subspace of V .

Theorem
Every vector space V has a spanning set.

Reason: Since there is no constraint on the size of a spanning set, we can
take S = V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 22 / 73

Spanning sets, row spaces, and column spaces

Theorem
If S is a non-empty subset of vectors in a vector space V , then Span(S) is
a subspace of V .

Theorem
Every vector space V has a spanning set.

Reason: Since there is no constraint on the size of a spanning set, we can
take S = V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 22 / 73

Spanning sets, row spaces, and column spaces

Theorem
If S is a non-empty subset of vectors in a vector space V , then Span(S) is
a subspace of V .

Theorem
Every vector space V has a spanning set.

Reason: Since there is no constraint on the size of a spanning set, we can
take S = V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 22 / 73

Spanning sets, row spaces, and column spaces

Theorem
If S is a non-empty subset of vectors in a vector space V , then Span(S) is
a subspace of V .

Theorem
Every vector space V has a spanning set.

Reason: Since there is no constraint on the size of a spanning set, we can
take S = V .

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 22 / 73

Spanning sets, row spaces, and column spaces

Examples
In Rn we write ei for the n × 1 column vector which is zero in each entry
except the ith, where it is equal to 1 (in other words, in matrix notation,
ei (i , 1) = 1; ei (j , 1) = 0, j 6= i).

Then the set {e1, e2, . . . , en} spans Rn, since for any vector
v = [a1, a2, . . . , an]T we have

[a1, a2, . . . , an]T =
n∑

i=1
aiei

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 23 / 73

Spanning sets, row spaces, and column spaces

Examples
In Rn we write ei for the n × 1 column vector which is zero in each entry
except the ith, where it is equal to 1 (in other words, in matrix notation,
ei (i , 1) = 1; ei (j , 1) = 0, j 6= i).

Then the set {e1, e2, . . . , en} spans Rn, since for any vector
v = [a1, a2, . . . , an]T we have

[a1, a2, . . . , an]T =
n∑

i=1
aiei

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 23 / 73

Spanning sets, row spaces, and column spaces

Examples
In Rn we write ei for the n × 1 column vector which is zero in each entry
except the ith, where it is equal to 1 (in other words, in matrix notation,
ei (i , 1) = 1; ei (j , 1) = 0, j 6= i).

Then the set {e1, e2, . . . , en} spans Rn, since for any vector
v = [a1, a2, . . . , an]T we have

[a1, a2, . . . , an]T =
n∑

i=1
aiei

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 23 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition

S is a minimal spanning set for V if
V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition

S is a minimal spanning set for V if
V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition

S is a minimal spanning set for V if
V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition

S is a minimal spanning set for V if
V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition
S is a minimal spanning set for V if

V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition
S is a minimal spanning set for V if

V = Span(S), and

For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition
S is a minimal spanning set for V if

V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces

The previous discussion suggests that spanning sets are not only not
unique, they can have vastly different sizes.

For example, R2 is spanned by {e1, e2}. It is also spanned by the set R2
itself, which is much larger.

It is natural to ask how small a spanning set can be. This leads to

Minimal spanning set – definition
S is a minimal spanning set for V if

V = Span(S), and
For any proper subset T ( S, Span(T ) ( V .

Theorem
For all n ≥ 1 the set Sn := {e1, . . . , en} is a minimal spanning set for Rn.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 24 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.

If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition

If A is an m × n matrix the
row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm.

This leads to the following definition

Row and Column Space – Definition

If A is an m × n matrix the
row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition

If A is an m × n matrix the
row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition

If A is an m × n matrix the
row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition
If A is an m × n matrix the

row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition
If A is an m × n matrix the

row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the

column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition
If A is an m × n matrix the

row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces
Recall that for a matrix A we write A(i , 🙂 for the ith row, and A(:, j) for
the jth column.
If A is an m × n matrix, then each row A(i , 🙂 is a row vector in the vector
space R1×n, while each column is a vector in the vector space
Rm×1 = Rm. This leads to the following definition

Row and Column Space – Definition
If A is an m × n matrix the

row space of A is the subspace
R(A) := Span{A(i , :)}1≤i≤m ⊂ R1×n, while the
column space of A is the subspace
C(A) := Span{A(:, j)}1≤i≤n ⊂ Rm×1.

Note that the row and column spaces of a matrix are different vector
spaces, unless m = n = 1. If m = n > 1 then R1×n and Rn×1 can be
naturally identified via the transpose operation, but they are not the same
space.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 25 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem

If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same). Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem

If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same). Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem

If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same). Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem
If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same).

Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem
If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same). Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Finding spanning sets for R(A) or C(A) is not a problem; they are just the
set of rows and columns of A.

However if one wants to find a minimal spanning set then some work
needs to be done. The main tool we will use is the following theorem.

Theorem
If A and B are matrices with A row equivalent to B then R(A) = R(B)
(their row spaces are the same). Moreover linear relations among the
columns of A and B are the same. This applies in particular if
B = rref (A).

We also have the following useful result.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 26 / 73

Spanning sets, row spaces, and column spaces

Theorem

If A is in reduced row echelon form, then
the columns of A containing leading ones are a linearly independent
set of column vectors.

Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);

the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors.

Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);

the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors.

Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);
the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors. Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.

Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);
the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors. Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);

the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors. Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);
the rows of A containing leading ones are a linearly independent set of
row vectors.

As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

Theorem
If A is in reduced row echelon form, then

the columns of A containing leading ones are a linearly independent
set of column vectors. Moreover, the columns that don’t contain
leading ones can be written as linear combination of the ones that do.
Consequently, the columns of A which contain leading ones form a
minimal spanning set for the column space C(A);
the rows of A containing leading ones are a linearly independent set of
row vectors. As all remaining rows must be identically zero, the rows
of A which contain leading ones form a minimal spanning set for the
row space R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 27 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by
computing rref (A);
identifying the set D = {i1, . . . , ik} of pivot columns;
choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by

computing rref (A);
identifying the set D = {i1, . . . , ik} of pivot columns;
choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by
computing rref (A);

identifying the set D = {i1, . . . , ik} of pivot columns;
choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by
computing rref (A);
identifying the set D = {i1, . . . , ik} of pivot columns;

choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by
computing rref (A);
identifying the set D = {i1, . . . , ik} of pivot columns;
choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

This theorem provides an efficient algorithm for determining minimal
spanning sets, as well as determining linear relations among vectors.

Given a matrix A, a minimal spanning set for C(A) is found by
computing rref (A);
identifying the set D = {i1, . . . , ik} of pivot columns;
choosing your minimal spanning set for C(A) to be the subset of
column vectors

{A(:, i1), . . . ,A(:, ik)}

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 28 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.

And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).

Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).

Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);

identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).

Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);

choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).

Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).
Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).
Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).

Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Moreover, in the reduced row echelon form the columns that are not pivot
columns are easily computed as linear combinations of the pivot columns.
And since the row space remains unchanged under row operations, a
minimal spanning set for R(A) = R(rref (A)) is found by

computing rref (A);
identifying the set D = {1, . . . , k} of indices corresponding to the
non-zero rows of rref (A);
choosing as your minimal spanning set for R(A) the subset of row
vectors

{B(1, :), . . . ,B(k, :)}

where B = rref (A).
Note that, unlike the column case, these rows typically won’t be among
the original set of rows of A; they will simply have the same span.

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 29 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;

B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);

B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).
So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples
Suppose A is a 4× 5 matrix with

B = rref (A) =




1 0 3 0 4
0 1 −2 0 5
0 0 0 1 7
0 0 0 0 0




We don’t know what the original matrix is, but for B = rref (A) we see
that

columns 1,2, and 4 of B are linearly independent;
B(:, 3) = 3B(:, 1)− 2B(:, 2);
B(:, 5) = 4B(:, 1) + 5B(:, 2) + 7B(:, 4).

So we can conclude the same must be true for A:

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 30 / 73

Spanning sets, row spaces, and column spaces

Examples

columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;

A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);

A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore

*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while

*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

Examples
columns 1,2, and 4 of A are linearly independent;
A(:, 3) = 3A(:, 1)− 2A(:, 2);
A(:, 5) = 4A(:, 1) + 5A(:, 2) + 7A(:, 4).

Therefore
*) {A(:, 1),A(:, 2),A(:, 4)} form a minimal spanning set for the columns
space C(A), while
*) the rows B(1, :),B(2, :),B(3, 🙂 form a minimal spanning set for the row
space R(B) = R(A).

Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 31 / 73

Spanning sets, row spaces, and column spaces

There are additional implications of the above.

Additional results

If A is an m × n matrix, then
the columns of A are linearly independent precisely when every
column of rref (A) contains a leading 1 (is a pivot column);
the columns of A span Rm (that is, C(A) = Rm) precisely when each
row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence
if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;

if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Spanning sets, row spaces, and column spaces There are additional implications of the above. Additional results If A is an m × n matrix, then the columns of A are linearly independent precisely when every column of rref (A) contains a leading 1 (is a pivot column); the columns of A span Rm (that is, C(A) = Rm) precisely when each row of rref (A) contains a leading 1 (equivalently, is non-zero). Hence if A is m × n with m > n then the columns may be linearly
independent, but they cannot span all of Rm;
if A is m × n with m < n then the columns may span all of Rm but cannot be linearly independent; if A is m × n with m = n then C(A) = Rm if and only if the columns of A are linearly independent, if and only if the columns of A are a basis for Rm. Moreover this happens precisely when rref (A) = Id . Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 32 / 73 Nullspaces If A is an m × n matrix with entries in R, the nullspace of A is the set N(A) := {x ∈ Rn | A ∗ x = 0} ⊂ Rn In other words, N(A) is the set of all solutions to the homogeneous matrix equation A ∗ x = 0. Theorem For any m × n matrix A with entries in R, N(A) is a subspace of Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 33 / 73 Nullspaces If A is an m × n matrix with entries in R, the nullspace of A is the set N(A) := {x ∈ Rn | A ∗ x = 0} ⊂ Rn In other words, N(A) is the set of all solutions to the homogeneous matrix equation A ∗ x = 0. Theorem For any m × n matrix A with entries in R, N(A) is a subspace of Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 33 / 73 Nullspaces If A is an m × n matrix with entries in R, the nullspace of A is the set N(A) := {x ∈ Rn | A ∗ x = 0} ⊂ Rn In other words, N(A) is the set of all solutions to the homogeneous matrix equation A ∗ x = 0. Theorem For any m × n matrix A with entries in R, N(A) is a subspace of Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 33 / 73 Nullspaces If A is an m × n matrix with entries in R, the nullspace of A is the set N(A) := {x ∈ Rn | A ∗ x = 0} ⊂ Rn In other words, N(A) is the set of all solutions to the homogeneous matrix equation A ∗ x = 0. Theorem For any m × n matrix A with entries in R, N(A) is a subspace of Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 33 / 73 Nullspaces Nullspaces can be computed using methods that have already been discussed. If A is an m × n matrix the equation A ∗ x = 0 is consistent. Either the equation only admits the trivial solution x = 0, or we have a parametrized family of solutions. In either case we want to find a spanning set for N(A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 34 / 73 Nullspaces Nullspaces can be computed using methods that have already been discussed. If A is an m × n matrix the equation A ∗ x = 0 is consistent. Either the equation only admits the trivial solution x = 0, or we have a parametrized family of solutions. In either case we want to find a spanning set for N(A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 34 / 73 Nullspaces Nullspaces can be computed using methods that have already been discussed. If A is an m × n matrix the equation A ∗ x = 0 is consistent. Either the equation only admits the trivial solution x = 0, or we have a parametrized family of solutions. In either case we want to find a spanning set for N(A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 34 / 73 Nullspaces Examples Suppose A is a 5× 5 matrix for which rref (A) =   1 0 0 −1 2 0 1 0 2 3 0 0 1 −4 −3 0 0 0 0 0 0 0 0 0 0   The columns which do not contain leading ones are the fourth and fifth. Thus x4 and x5 are the natural parameters for the parametrized solution set. We can derive a spanning set for N(A) as follows: Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 35 / 73 Nullspaces Examples Suppose A is a 5× 5 matrix for which rref (A) =   1 0 0 −1 2 0 1 0 2 3 0 0 1 −4 −3 0 0 0 0 0 0 0 0 0 0   The columns which do not contain leading ones are the fourth and fifth. Thus x4 and x5 are the natural parameters for the parametrized solution set. We can derive a spanning set for N(A) as follows: Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 35 / 73 Nullspaces Examples Suppose A is a 5× 5 matrix for which rref (A) =   1 0 0 −1 2 0 1 0 2 3 0 0 1 −4 −3 0 0 0 0 0 0 0 0 0 0   The columns which do not contain leading ones are the fourth and fifth. Thus x4 and x5 are the natural parameters for the parametrized solution set. We can derive a spanning set for N(A) as follows: Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 35 / 73 Nullspaces Examples Suppose A is a 5× 5 matrix for which rref (A) =   1 0 0 −1 2 0 1 0 2 3 0 0 1 −4 −3 0 0 0 0 0 0 0 0 0 0   The columns which do not contain leading ones are the fourth and fifth. Thus x4 and x5 are the natural parameters for the parametrized solution set. We can derive a spanning set for N(A) as follows: Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 35 / 73 Nullspaces Examples [Step 1] Write the all of the original set of variables x1, x2, x3, x4, x5 in terms of the free variables x4, x5: x1 = x4 − 2x5 x2 = −2x4 − 3x5 x3 = 4x4 + 3x5 x4 = x4 x5 = x5 Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 36 / 73 Nullspaces Examples [Step 2] Use this to write the general solution vector x in terms of x4 and x5: x =   x1 x2 x3 x4 x5   =   x4 − 2x5 −2x4 − 3x5 4x4 + 3x5 x4 x5   [Step 3] Separate the general solution into its homogeneous components: x =   x4 − 2x5 −2x4 − 3x5 4x4 + 3x5 x4 x5   =   x4 −2x4 4x4 x4 0  +   −2x5 −3x5 3x5 0 x5   = x4   1 −2 4 1 0  + x5   −2 −3 3 0 1   Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 37 / 73 Nullspaces Examples [Step 2] Use this to write the general solution vector x in terms of x4 and x5: x =   x1 x2 x3 x4 x5   =   x4 − 2x5 −2x4 − 3x5 4x4 + 3x5 x4 x5   [Step 3] Separate the general solution into its homogeneous components: x =   x4 − 2x5 −2x4 − 3x5 4x4 + 3x5 x4 x5   =   x4 −2x4 4x4 x4 0  +   −2x5 −3x5 3x5 0 x5   = x4   1 −2 4 1 0  + x5   −2 −3 3 0 1   Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 37 / 73 Nullspaces Examples [Step 4] Use this last expression for the general solution to finally express N(A) as the span of a set of vectors: N(A) = Span     1 −2 4 1 0   ,   −2 −3 3 0 1     Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 38 / 73 Range The range of an m × n matrix A is the subspace Range(A) := {y ∈ Rm | A ∗ x = y is consistent} ⊆ Rm The Consistency Theorem tells us that A ∗ x = y is consistent precisely when y is a linear combination of the columns of A. Consequently Theorem For all real matrices A, Range(A) = C(A) (the column space of A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 39 / 73 Range The range of an m × n matrix A is the subspace Range(A) := {y ∈ Rm | A ∗ x = y is consistent} ⊆ Rm The Consistency Theorem tells us that A ∗ x = y is consistent precisely when y is a linear combination of the columns of A. Consequently Theorem For all real matrices A, Range(A) = C(A) (the column space of A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 39 / 73 Range The range of an m × n matrix A is the subspace Range(A) := {y ∈ Rm | A ∗ x = y is consistent} ⊆ Rm The Consistency Theorem tells us that A ∗ x = y is consistent precisely when y is a linear combination of the columns of A. Consequently Theorem For all real matrices A, Range(A) = C(A) (the column space of A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 39 / 73 Range The range of an m × n matrix A is the subspace Range(A) := {y ∈ Rm | A ∗ x = y is consistent} ⊆ Rm The Consistency Theorem tells us that A ∗ x = y is consistent precisely when y is a linear combination of the columns of A. Consequently Theorem For all real matrices A, Range(A) = C(A) (the column space of A). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 39 / 73 Bases and Dimension Basis - Definition A set of vectors S = {v1, . . . , vn} ⊂ V is a basis for V if it spans V and it is linearly independent. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 40 / 73 Bases and Dimension Basis - Definition A set of vectors S = {v1, . . . , vn} ⊂ V is a basis for V if it spans V and it is linearly independent. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 40 / 73 Bases and Dimension Basis - Definition A set of vectors S = {v1, . . . , vn} ⊂ V is a basis for V if it spans V and it is linearly independent. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 40 / 73 Bases and Dimension Basis - Definition A set of vectors S = {v1, . . . , vn} ⊂ V is a basis for V if it spans V and it is linearly independent. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 40 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Fundamental Properties [B1] A non-zero set S = {v1, . . . , vn} ⊂ V is a basis for V iff it is a minimal spanning set for V . [B2] Every non-zero vector space V admits a basis. [B3] (finite case) If S = {v1, . . . , vn} and T = {w1, . . . ,wm} are two bases for V 6= {0}, then m = n. Dimension - Definition If V is a non-zero vector space, the dimension of V is the number of vectors in a basis for V : Dim(V ) := #S where S is a basis for V (If V = {0} we set Dim(V ) := 0). Property B3 indicates that the dimension of a (non-zero) vector space V doesn’t depend on the choice of basis for V ; it only depends on V . The dimension of a vector space is the single most important numerical invariant one can attach to that space. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 41 / 73 Bases and Dimension Suppose W is a subspace of Rn, and we wish to find a basis. The first step is to determine a spanning set S = {w1, . . . ,wm} in the event it is not already given. Then forming the matrix A = [ w1 w2 . . . wm ] we observe that (by construction) W = C(A), the column space of A. We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set T ⊆ S from among the original vectors in S. Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of Rn (for some n). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 42 / 73 Bases and Dimension Suppose W is a subspace of Rn, and we wish to find a basis. The first step is to determine a spanning set S = {w1, . . . ,wm} in the event it is not already given. Then forming the matrix A = [ w1 w2 . . . wm ] we observe that (by construction) W = C(A), the column space of A. We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set T ⊆ S from among the original vectors in S. Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of Rn (for some n). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 42 / 73 Bases and Dimension Suppose W is a subspace of Rn, and we wish to find a basis. The first step is to determine a spanning set S = {w1, . . . ,wm} in the event it is not already given. Then forming the matrix A = [ w1 w2 . . . wm ] we observe that (by construction) W = C(A), the column space of A. We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set T ⊆ S from among the original vectors in S. Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of Rn (for some n). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 42 / 73 Bases and Dimension Suppose W is a subspace of Rn, and we wish to find a basis. The first step is to determine a spanning set S = {w1, . . . ,wm} in the event it is not already given. Then forming the matrix A = [ w1 w2 . . . wm ] we observe that (by construction) W = C(A), the column space of A. We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set T ⊆ S from among the original vectors in S. Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of Rn (for some n). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 42 / 73 Bases and Dimension Suppose W is a subspace of Rn, and we wish to find a basis. The first step is to determine a spanning set S = {w1, . . . ,wm} in the event it is not already given. Then forming the matrix A = [ w1 w2 . . . wm ] we observe that (by construction) W = C(A), the column space of A. We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set T ⊆ S from among the original vectors in S. Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of Rn (for some n). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 42 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension An important consequence of the above is Theorem If V is a finite-dimensional vector space and W ⊂ V is a subspace, then Dim(W ) ≤ Dim(V ); moreover Dim(W ) = Dim(V ) iff W = V . We will occasionally write ei ∈ Rn as eni when we want to emphasize the fact that the vector lies in Rn. Given the equivalence between bases and minimal spanning sets, we have Theorem For all n ≥ 1, the set e := {en1, . . . , e n n} is a basis for Rn. The basis e is referred to as the standard basis for Rn. The above method works for finding a basis for a subspace of Rn. In the event the finite-dimensional vector space W is not naturally a subspace of Rn for any n this method still applies with some additional work. This is discussed below. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 43 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Bases and Dimension Rank and Nullity - Definition For an m × n real matrix A, the rank of A is the dimension of the column space of A - Rank(A) := Dim(C(A)); nullity of A is the dimension of the nullspace of A - Null(A) := Dim(N(A)). From above we see that Rank(A) = the number of columns of rref (A) that contain a leading 1, while Null(A) = the number of columns of rref (A) that do not contain a leading 1. Since the number of columns of rref (A) is the same as the number of columns of A we have Rank - Nullity Theorem For any real m × n matrix A, N(A) + R(A) = n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 44 / 73 Coordinate Systems From the beginning we have adopted the convention that vectors in Rn are represented by n × 1 matrices - column vectors - with entries in R. Another, equivalent way to arrive at such a description is to start with the standard basis e := {en1, e n 2, . . . , e n n}, viewed simply as a linearly independent set, and then define Rn to be the span of these basis vectors. In this way, we realize that when writing down an n × 1 column vector representing v, we are simply recording (in columnar form) the coefficients that occur when representing v as a linear combination of standard basis vectors: v = [a1 a2 . . . an]T ⇔ v = a1en1 + a2e n 2 + · · ·+ ane n n We notice now that this can be done not just for the standard basis in Rn, but for any basis in any vector space. For the purposes of this section and what follows, we will only be concerned with such representations in finite dimensional vector spaces (or subspaces). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 45 / 73 Coordinate Systems From the beginning we have adopted the convention that vectors in Rn are represented by n × 1 matrices - column vectors - with entries in R. Another, equivalent way to arrive at such a description is to start with the standard basis e := {en1, e n 2, . . . , e n n}, viewed simply as a linearly independent set, and then define Rn to be the span of these basis vectors. In this way, we realize that when writing down an n × 1 column vector representing v, we are simply recording (in columnar form) the coefficients that occur when representing v as a linear combination of standard basis vectors: v = [a1 a2 . . . an]T ⇔ v = a1en1 + a2e n 2 + · · ·+ ane n n We notice now that this can be done not just for the standard basis in Rn, but for any basis in any vector space. For the purposes of this section and what follows, we will only be concerned with such representations in finite dimensional vector spaces (or subspaces). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 45 / 73 Coordinate Systems From the beginning we have adopted the convention that vectors in Rn are represented by n × 1 matrices - column vectors - with entries in R. Another, equivalent way to arrive at such a description is to start with the standard basis e := {en1, e n 2, . . . , e n n}, viewed simply as a linearly independent set, and then define Rn to be the span of these basis vectors. In this way, we realize that when writing down an n × 1 column vector representing v, we are simply recording (in columnar form) the coefficients that occur when representing v as a linear combination of standard basis vectors: v = [a1 a2 . . . an]T ⇔ v = a1en1 + a2e n 2 + · · ·+ ane n n We notice now that this can be done not just for the standard basis in Rn, but for any basis in any vector space. For the purposes of this section and what follows, we will only be concerned with such representations in finite dimensional vector spaces (or subspaces). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 45 / 73 Coordinate Systems From the beginning we have adopted the convention that vectors in Rn are represented by n × 1 matrices - column vectors - with entries in R. Another, equivalent way to arrive at such a description is to start with the standard basis e := {en1, e n 2, . . . , e n n}, viewed simply as a linearly independent set, and then define Rn to be the span of these basis vectors. In this way, we realize that when writing down an n × 1 column vector representing v, we are simply recording (in columnar form) the coefficients that occur when representing v as a linear combination of standard basis vectors: v = [a1 a2 . . . an]T ⇔ v = a1en1 + a2e n 2 + · · ·+ ane n n We notice now that this can be done not just for the standard basis in Rn, but for any basis in any vector space. For the purposes of this section and what follows, we will only be concerned with such representations in finite dimensional vector spaces (or subspaces). Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 45 / 73 Coordinate systems Coordinate vector with respect to a basis - Definition If b = {u1,u2, . . . ,un} is a basis for a (finite dimensional) vector space V , and v ∈ V , the coordinate vector bv of v with respect to the basis b is the n × 1 column vector which records the unique set of coefficients needed to represent v as a linear combination of the basis vectors in b: bv = [a1 a2 . . . an]T ⇔ v = a1u1 + a2u2 + · · ·+ anun Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 46 / 73 Coordinate systems Coordinate vector with respect to a basis - Definition If b = {u1,u2, . . . ,un} is a basis for a (finite dimensional) vector space V , and v ∈ V , the coordinate vector bv of v with respect to the basis b is the n × 1 column vector which records the unique set of coefficients needed to represent v as a linear combination of the basis vectors in b: bv = [a1 a2 . . . an]T ⇔ v = a1u1 + a2u2 + · · ·+ anun Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 46 / 73 Coordinate systems It is important here to distinguish between i) a vector v ∈ V and ii) its coordinate representation with respect to a particular basis for V . For that reason, vectors in Rn - written as n × 1 column vectors - may some times be written with the decoration indicating that we are really looking at the representation of the vector with respect to the standard basis. As we will see, this additional decoration provides necessary book-keeping in the event we are considering representations with respect to different bases in Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 47 / 73 Coordinate systems It is important here to distinguish between i) a vector v ∈ V and ii) its coordinate representation with respect to a particular basis for V . For that reason, vectors in Rn - written as n × 1 column vectors - may some times be written with the decoration indicating that we are really looking at the representation of the vector with respect to the standard basis. As we will see, this additional decoration provides necessary book-keeping in the event we are considering representations with respect to different bases in Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 47 / 73 Coordinate systems It is important here to distinguish between i) a vector v ∈ V and ii) its coordinate representation with respect to a particular basis for V . For that reason, vectors in Rn - written as n × 1 column vectors - may some times be written with the decoration indicating that we are really looking at the representation of the vector with respect to the standard basis. As we will see, this additional decoration provides necessary book-keeping in the event we are considering representations with respect to different bases in Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 47 / 73 Coordinate systems It is important here to distinguish between i) a vector v ∈ V and ii) its coordinate representation with respect to a particular basis for V . For that reason, vectors in Rn - written as n × 1 column vectors - may some times be written with the decoration indicating that we are really looking at the representation of the vector with respect to the standard basis. As we will see, this additional decoration provides necessary book-keeping in the event we are considering representations with respect to different bases in Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 47 / 73 Coordinate systems It is important here to distinguish between i) a vector v ∈ V and ii) its coordinate representation with respect to a particular basis for V . For that reason, vectors in Rn - written as n × 1 column vectors - may some times be written with the decoration indicating that we are really looking at the representation of the vector with respect to the standard basis. As we will see, this additional decoration provides necessary book-keeping in the event we are considering representations with respect to different bases in Rn. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 47 / 73 Coordinate systems Examples Let v ∈ R3 with ev = [2 3 − 1]T , and let b = {u1,u2,u3} be the basis of R3 given by eu1 = [1 1 0]T , eu2 = [1 0 1]T , eu3 = [0 1 1]T We wish to determine bv, the coordinate representation of v with respect to the basis b. In other words, solve for the coefficients in the vector equation v = x1u1 + x2u2 + x3u3 Referring back to the consistency theorem for systems of equations, we see that solving for x = [x1 x2 x3]T is equivalent to solving for x in the matrix equation A ∗ x = evT , where A is the 3× 3 matrix A = [ eu1 eu2 eu3 ] : Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 48 / 73 Coordinate systems Examples Let v ∈ R3 with ev = [2 3 − 1]T , and let b = {u1,u2,u3} be the basis of R3 given by eu1 = [1 1 0]T , eu2 = [1 0 1]T , eu3 = [0 1 1]T We wish to determine bv, the coordinate representation of v with respect to the basis b. In other words, solve for the coefficients in the vector equation v = x1u1 + x2u2 + x3u3 Referring back to the consistency theorem for systems of equations, we see that solving for x = [x1 x2 x3]T is equivalent to solving for x in the matrix equation A ∗ x = evT , where A is the 3× 3 matrix A = [ eu1 eu2 eu3 ] : Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 48 / 73 Coordinate systems Examples Let v ∈ R3 with ev = [2 3 − 1]T , and let b = {u1,u2,u3} be the basis of R3 given by eu1 = [1 1 0]T , eu2 = [1 0 1]T , eu3 = [0 1 1]T We wish to determine bv, the coordinate representation of v with respect to the basis b. In other words, solve for the coefficients in the vector equation v = x1u1 + x2u2 + x3u3 Referring back to the consistency theorem for systems of equations, we see that solving for x = [x1 x2 x3]T is equivalent to solving for x in the matrix equation A ∗ x = evT , where A is the 3× 3 matrix A = [ eu1 eu2 eu3 ] : Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 48 / 73 Coordinate systems Examples Let v ∈ R3 with ev = [2 3 − 1]T , and let b = {u1,u2,u3} be the basis of R3 given by eu1 = [1 1 0]T , eu2 = [1 0 1]T , eu3 = [0 1 1]T We wish to determine bv, the coordinate representation of v with respect to the basis b. In other words, solve for the coefficients in the vector equation v = x1u1 + x2u2 + x3u3 Referring back to the consistency theorem for systems of equations, we see that solving for x = [x1 x2 x3]T is equivalent to solving for x in the matrix equation A ∗ x = evT , where A is the 3× 3 matrix A = [ eu1 eu2 eu3 ] : Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 48 / 73 Coordinate systems Examples  1 1 01 0 1 0 1 1   ∗  x1x2 x3   =   23 −1   Forming the ACM and putting it into reduced row echelon form yields rref    1 1 0 21 0 1 3 0 1 1 −1     =  1 0 0 30 1 0 −1 0 0 1 0   from which we see that x = [3 − 1 0]T = bv Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 49 / 73 Coordinate systems Examples  1 1 01 0 1 0 1 1   ∗  x1x2 x3   =   23 −1   Forming the ACM and putting it into reduced row echelon form yields rref    1 1 0 21 0 1 3 0 1 1 −1     =  1 0 0 30 1 0 −1 0 0 1 0   from which we see that x = [3 − 1 0]T = bv Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 49 / 73 Coordinate systems Examples  1 1 01 0 1 0 1 1   ∗  x1x2 x3   =   23 −1   Forming the ACM and putting it into reduced row echelon form yields rref    1 1 0 21 0 1 3 0 1 1 −1     =  1 0 0 30 1 0 −1 0 0 1 0   from which we see that x = [3 − 1 0]T = bv Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 49 / 73 Coordinate systems Examples  1 1 01 0 1 0 1 1   ∗  x1x2 x3   =   23 −1   Forming the ACM and putting it into reduced row echelon form yields rref    1 1 0 21 0 1 3 0 1 1 −1     =  1 0 0 30 1 0 −1 0 0 1 0   from which we see that x = [3 − 1 0]T = bv Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 49 / 73 Coordinate systems This last example illustrates one of the basic computations we will want to be able to do; given the coordinate description of a vector with respect to one basis, find its coordinate representation with respect to a (specified) different basis. The matrix A in the above example is referred to as a transition matrix; specifically the base transition matrix from the basis b to the basis e, where b = {u1,u2,u3} As we will see in more detail below, the notation for the base transition matrix from b to e is eTb := [ eu1 eu2 eu3 ] For now we will simply define base transition matrices. The proof of the equality in equation (1) below will be given in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 50 / 73 Coordinate systems This last example illustrates one of the basic computations we will want to be able to do; given the coordinate description of a vector with respect to one basis, find its coordinate representation with respect to a (specified) different basis. The matrix A in the above example is referred to as a transition matrix; specifically the base transition matrix from the basis b to the basis e, where b = {u1,u2,u3} As we will see in more detail below, the notation for the base transition matrix from b to e is eTb := [ eu1 eu2 eu3 ] For now we will simply define base transition matrices. The proof of the equality in equation (1) below will be given in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 50 / 73 Coordinate systems This last example illustrates one of the basic computations we will want to be able to do; given the coordinate description of a vector with respect to one basis, find its coordinate representation with respect to a (specified) different basis. The matrix A in the above example is referred to as a transition matrix; specifically the base transition matrix from the basis b to the basis e, where b = {u1,u2,u3} As we will see in more detail below, the notation for the base transition matrix from b to e is eTb := [ eu1 eu2 eu3 ] For now we will simply define base transition matrices. The proof of the equality in equation (1) below will be given in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 50 / 73 Coordinate systems This last example illustrates one of the basic computations we will want to be able to do; given the coordinate description of a vector with respect to one basis, find its coordinate representation with respect to a (specified) different basis. The matrix A in the above example is referred to as a transition matrix; specifically the base transition matrix from the basis b to the basis e, where b = {u1,u2,u3} As we will see in more detail below, the notation for the base transition matrix from b to e is eTb := [ eu1 eu2 eu3 ] For now we will simply define base transition matrices. The proof of the equality in equation (1) below will be given in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 50 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems Base transition matrices for Rn - Definition If b = {u1,u2, . . . ,un} is a basis for Rn, then the base transition matrix from b to e is the n × n non-singular matrix eTb := [ eu1 eu2 . . . eun ] where eui is the coordinate vector of ui with respect to the standard basis e of Rn. The base transition matrix bTe from e to b is then given as the inverse of eTb bTe := eT−1b Finally, if ev is the coordinate representation of v ∈ Rn with respect to the standard basis, then the coordinate representation of v with respect to the basis b, written as Sv, can be computed as a product bv = bTe ∗ ev (1) Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 51 / 73 Coordinate systems The discussion of base change in Rn applies more generally to transitioning between two bases for an arbitrary vector space V . The setup for this (and the proof that it works as claimed) is best handled within the more general framework of linear transformations and their matrix representations, which we discuss in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 52 / 73 Coordinate systems The discussion of base change in Rn applies more generally to transitioning between two bases for an arbitrary vector space V . The setup for this (and the proof that it works as claimed) is best handled within the more general framework of linear transformations and their matrix representations, which we discuss in the next section. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 52 / 73 Table of Contents 1 Vector Spaces The space Rn Vector Spaces - Definition Subspaces Spanning Sets, Row Spaces, Column Spaces Nullspaces Range Bases and Dimension Coordinate Systems 2 Linear Transformations Linear Transformations - Definition, properties, and terminology Linear Transformations - Matrix representation Change of basis Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 53 / 73 Linear Transformations - Definition Suppose V and W are vector spaces (of arbitrary dimension). A function f : V →W is (as we recall) a rule which associates to each v ∈ V a unique vector f (v) ∈W . We will call such a function a linear transformation if it commutes with the linear structure in the domain and range: f (v + w) = f (v) + f (w) ∀v,w ∈ V f (αv) = αf (v) ∀α ∈ R, v ∈ V In other words, f takes sums to sums and scalar products to scalar products. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 54 / 73 Linear Transformations - Definition Suppose V and W are vector spaces (of arbitrary dimension). A function f : V →W is (as we recall) a rule which associates to each v ∈ V a unique vector f (v) ∈W . We will call such a function a linear transformation if it commutes with the linear structure in the domain and range: f (v + w) = f (v) + f (w) ∀v,w ∈ V f (αv) = αf (v) ∀α ∈ R, v ∈ V In other words, f takes sums to sums and scalar products to scalar products. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 54 / 73 Linear Transformations - Definition Suppose V and W are vector spaces (of arbitrary dimension). A function f : V →W is (as we recall) a rule which associates to each v ∈ V a unique vector f (v) ∈W . We will call such a function a linear transformation if it commutes with the linear structure in the domain and range: f (v + w) = f (v) + f (w) ∀v,w ∈ V f (αv) = αf (v) ∀α ∈ R, v ∈ V In other words, f takes sums to sums and scalar products to scalar products. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 54 / 73 Linear Transformations - Definition These two properties can be combined into one, using linear combinations. Precisely, Definition Let V ,W be vector spaces. A function f : V →W is a linear transformation if f (αv + βw) = αf (v) + βf (w) ∀α, β ∈ R, v,w ∈ V Theorem A map f : V →W is a linear transformation iff f (α1v1 + . . . αnvn) = α1f (v1) + . . . αnf (vn) ∀αi ∈ R, vi ∈ V Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 55 / 73 Linear Transformations - Definition These two properties can be combined into one, using linear combinations. Precisely, Definition Let V ,W be vector spaces. A function f : V →W is a linear transformation if f (αv + βw) = αf (v) + βf (w) ∀α, β ∈ R, v,w ∈ V Theorem A map f : V →W is a linear transformation iff f (α1v1 + . . . αnvn) = α1f (v1) + . . . αnf (vn) ∀αi ∈ R, vi ∈ V Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 55 / 73 Linear Transformations - Definition These two properties can be combined into one, using linear combinations. Precisely, Definition Let V ,W be vector spaces. A function f : V →W is a linear transformation if f (αv + βw) = αf (v) + βf (w) ∀α, β ∈ R, v,w ∈ V Theorem A map f : V →W is a linear transformation iff f (α1v1 + . . . αnvn) = α1f (v1) + . . . αnf (vn) ∀αi ∈ R, vi ∈ V Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 55 / 73 Linear Transformations - Definition These two properties can be combined into one, using linear combinations. Precisely, Definition Let V ,W be vector spaces. A function f : V →W is a linear transformation if f (αv + βw) = αf (v) + βf (w) ∀α, β ∈ R, v,w ∈ V Theorem A map f : V →W is a linear transformation iff f (α1v1 + . . . αnvn) = α1f (v1) + . . . αnf (vn) ∀αi ∈ R, vi ∈ V Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 55 / 73 Linear Transformations - Definition These two properties can be combined into one, using linear combinations. Precisely, Definition Let V ,W be vector spaces. A function f : V →W is a linear transformation if f (αv + βw) = αf (v) + βf (w) ∀α, β ∈ R, v,w ∈ V Theorem A map f : V →W is a linear transformation iff f (α1v1 + . . . αnvn) = α1f (v1) + . . . αnf (vn) ∀αi ∈ R, vi ∈ V Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 55 / 73 Linear Transformations - Properties As a consequence of this, we have Theorem If f : V →W is a linear transformation, then it is uniquely determined by its values on a basis for V . Conversely, if S := {v1, . . . vn} is a basis for V and g : S →W is a function from the set S to W , then g extends uniquely to a linear transformation f : V →W with f (vi ) = g(vi ), 1 ≤ i ≤ n. Thus, if f1, f2 : V →W are two linear transformations which agree on a basis for V , then f1 = f2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 56 / 73 Linear Transformations - Properties As a consequence of this, we have Theorem If f : V →W is a linear transformation, then it is uniquely determined by its values on a basis for V . Conversely, if S := {v1, . . . vn} is a basis for V and g : S →W is a function from the set S to W , then g extends uniquely to a linear transformation f : V →W with f (vi ) = g(vi ), 1 ≤ i ≤ n. Thus, if f1, f2 : V →W are two linear transformations which agree on a basis for V , then f1 = f2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 56 / 73 Linear Transformations - Properties As a consequence of this, we have Theorem If f : V →W is a linear transformation, then it is uniquely determined by its values on a basis for V . Conversely, if S := {v1, . . . vn} is a basis for V and g : S →W is a function from the set S to W , then g extends uniquely to a linear transformation f : V →W with f (vi ) = g(vi ), 1 ≤ i ≤ n. Thus, if f1, f2 : V →W are two linear transformations which agree on a basis for V , then f1 = f2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 56 / 73 Linear Transformations - Properties As a consequence of this, we have Theorem If f : V →W is a linear transformation, then it is uniquely determined by its values on a basis for V . Conversely, if S := {v1, . . . vn} is a basis for V and g : S →W is a function from the set S to W , then g extends uniquely to a linear transformation f : V →W with f (vi ) = g(vi ), 1 ≤ i ≤ n. Thus, if f1, f2 : V →W are two linear transformations which agree on a basis for V , then f1 = f2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 56 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Kernel and Image Linear transformations may be used to define certain subspaces. Let L : V →W be a linear transformation. The kernel of L is then ker(L) := {v ∈ V | L(v) = 0} ⊂ V The image of L is defined as im(L) := {w ∈W | ∃v ∈ V with L(v) = w} ⊂W The image of L is sometimes denoted L(V ). It is also referred to as the range of L. These subspaces are useful in defining specific types of linear transformations. Monomorphims, epimorphisms, and isomorphisms L : V →W is surjective or an epimorphism iff im(L) = W ; injective or a monomorphism iff ker(L) = {0}; bijective or an isomorphism iff L is both surjective and injective. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 57 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Isomorphisms Two vector spaces V and W are isomorphic - denoted V ∼= W - if there exists a linear transformation L : V →W which is an isomorphism. Theorem For any linear transformation L : V →W , ker(L) is a subspace of V and im(L) is a subspace of W . Moreover, Dim(V ) = Dim(ker(L)) + Dim(im(L)) If V and W are finite-dimensional, then V ∼= W iff Dim(V ) = Dim(W ). It is natural to ask: what does it really mean for a map to be a linear transformation? For example, if f : Rn → Rm, or more generally if f : V →W satisfies the above property, does it admit some simple description? We investigate this question next. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 58 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation A linear transformation can be represented in terms of multiplication by a matrix. Suppose V = Rn,W = Rm, and LA : V →W is given by LA(v) = A ∗ v for some m × n real matrix A. Then it follows immediately from the properties of matrix algebra that LA is a linear transformation: LA(αv + βw) = A ∗ (αv + βw) = α(A ∗ v) + β(A ∗w) = αLA(v) + βLA(w) Conversely, suppose the linear transformation L is given. Define the matrix AL by AL = [L(e1) L(e2) . . . L(en)] that is, the m × n matrix with A(:, i) = L(ei ), 1 ≤ i ≤ n. Then by construction AL ∗ (ei ) = A(:, i) = L(ei ), 1 ≤ i ≤ n Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 59 / 73 Linear Transformations - Matrix representation so that v 7→ L(v) and v 7→ AL ∗ v are two linear transformations which agree on a basis for Rn, which by the above implies L(v) = AL ∗ (v) ∀v ∈ Rn Because of this, the matrix AL is referred to as a matrix representation of L. Note that this representation is with respect to the standard basis for Rn and Rm. We see now that the same type of representation applies for arbitrary vector spaces once a basis has been fixed for both the domain and target. In other words, given Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 60 / 73 Linear Transformations - Matrix representation so that v 7→ L(v) and v 7→ AL ∗ v are two linear transformations which agree on a basis for Rn, which by the above implies L(v) = AL ∗ (v) ∀v ∈ Rn Because of this, the matrix AL is referred to as a matrix representation of L. Note that this representation is with respect to the standard basis for Rn and Rm. We see now that the same type of representation applies for arbitrary vector spaces once a basis has been fixed for both the domain and target. In other words, given Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 60 / 73 Linear Transformations - Matrix representation so that v 7→ L(v) and v 7→ AL ∗ v are two linear transformations which agree on a basis for Rn, which by the above implies L(v) = AL ∗ (v) ∀v ∈ Rn Because of this, the matrix AL is referred to as a matrix representation of L. Note that this representation is with respect to the standard basis for Rn and Rm. We see now that the same type of representation applies for arbitrary vector spaces once a basis has been fixed for both the domain and target. In other words, given Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 60 / 73 Linear Transformations - Matrix representation so that v 7→ L(v) and v 7→ AL ∗ v are two linear transformations which agree on a basis for Rn, which by the above implies L(v) = AL ∗ (v) ∀v ∈ Rn Because of this, the matrix AL is referred to as a matrix representation of L. Note that this representation is with respect to the standard basis for Rn and Rm. We see now that the same type of representation applies for arbitrary vector spaces once a basis has been fixed for both the domain and target. In other words, given Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 60 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation A vector space V with basis S = {v1, . . . , vn}, a vector space W with basis T = {w1, . . . ,wm}, and a linear transformation L : V →W we could ask if there is a similar representation of L in terms of a matrix (which depends on these two choices of bases). The answer is “yes”. Theorem For any v ∈ V T L(v) = T LS ∗ Sv where T LS is the m × n matrix defined by T LS = [T L(v1) T L(v2) . . . T L(vn)] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 61 / 73 Linear Transformations - Matrix representation Examples Suppose L : R3 → R2 is the linear transformation given by L(x) = L    x1x2 x3     = [ 2x1 − 3x2 + 7x3 4x2 − 5x3 ] To compute the matrix representation of L with respect to the standard bases of R3 and R2, we evaluate L on the basis vectors ei , 1 ≤ i ≤ 3 and then form the matrix AL = [ L(e1) L(e2) L(e3) ] , yielding AL = [ 2 −3 7 0 4 −5 ] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 62 / 73 Linear Transformations - Matrix representation Examples Suppose L : R3 → R2 is the linear transformation given by L(x) = L    x1x2 x3     = [ 2x1 − 3x2 + 7x3 4x2 − 5x3 ] To compute the matrix representation of L with respect to the standard bases of R3 and R2, we evaluate L on the basis vectors ei , 1 ≤ i ≤ 3 and then form the matrix AL = [ L(e1) L(e2) L(e3) ] , yielding AL = [ 2 −3 7 0 4 −5 ] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 62 / 73 Linear Transformations - Matrix representation Examples Suppose L : R3 → R2 is the linear transformation given by L(x) = L    x1x2 x3     = [ 2x1 − 3x2 + 7x3 4x2 − 5x3 ] To compute the matrix representation of L with respect to the standard bases of R3 and R2, we evaluate L on the basis vectors ei , 1 ≤ i ≤ 3 and then form the matrix AL = [ L(e1) L(e2) L(e3) ] , yielding AL = [ 2 −3 7 0 4 −5 ] Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 62 / 73 Linear Transformations - Matrix representation Examples Recall that Pn is the vector space of polynomials in one variable with real coefficients of degree less than 3. Let L : P3 → P3 be given by L(p(x)) = 2p′(x)− p(x), where p′(x) denotes the derivative of p(x). To find the matrix representation of L with respect to the basis S = {1, x , x2} for P3, we first compute L(1) = −1; L(x) = 2− 2x ; L(x2) = 4x − x2. This then yields SLS = [ SL(1) SL(x) SL(x2) ] =  −1 2 00 −2 4 0 0 −1   Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 63 / 73 Linear Transformations - Matrix representation Examples Recall that Pn is the vector space of polynomials in one variable with real coefficients of degree less than 3. Let L : P3 → P3 be given by L(p(x)) = 2p′(x)− p(x), where p′(x) denotes the derivative of p(x). To find the matrix representation of L with respect to the basis S = {1, x , x2} for P3, we first compute L(1) = −1; L(x) = 2− 2x ; L(x2) = 4x − x2. This then yields SLS = [ SL(1) SL(x) SL(x2) ] =  −1 2 00 −2 4 0 0 −1   Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 63 / 73 Linear Transformations - Matrix representation Examples Recall that Pn is the vector space of polynomials in one variable with real coefficients of degree less than 3. Let L : P3 → P3 be given by L(p(x)) = 2p′(x)− p(x), where p′(x) denotes the derivative of p(x). To find the matrix representation of L with respect to the basis S = {1, x , x2} for P3, we first compute L(1) = −1; L(x) = 2− 2x ; L(x2) = 4x − x2. This then yields SLS = [ SL(1) SL(x) SL(x2) ] =  −1 2 00 −2 4 0 0 −1   Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 63 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Returning once more to the general case where L : V →W is linear, S a basis for V , T a basis for W , we note that the bases S and T can be used to identify the kernel and image of L. Precisely, we have Theorem Assume Dim(V ) = n, Dim(W ) = m. Let A = T LS be the m × n matrix representation of L with respect to the pair of bases S,T . Then v ∈ ker(L) ⊂ V iff Sv ∈ N(A) ⊂ Rn. w ∈ im(L) ⊂W iff T w ∈ C(A) ⊂ Rm. This theorem tells us that, once we have fixed a basis for V and W , the representation of L by the matrix A = T LS further identifies i) the kernel ker(L) of L with the nullspace N(A) of A and ii) the image im(L) with the column space C(A) of A. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 64 / 73 Linear Transformations - Matrix representation Examples Suppose L : P3 → P3 is the linear transformation represented with respect to the standard basis on P3 by the matrix SLS = A =  2 3 13 9 6 1 6 5  . Our objective is to find a minimal spanning set for ker(L) and im(L) (with respect to the standard basis for P3). First we compute rref (A) =  1 0 −10 1 1 0 0 0  . We then see Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 65 / 73 Linear Transformations - Matrix representation Examples Suppose L : P3 → P3 is the linear transformation represented with respect to the standard basis on P3 by the matrix SLS = A =  2 3 13 9 6 1 6 5  . Our objective is to find a minimal spanning set for ker(L) and im(L) (with respect to the standard basis for P3). First we compute rref (A) =  1 0 −10 1 1 0 0 0  . We then see Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 65 / 73 Linear Transformations - Matrix representation Examples Suppose L : P3 → P3 is the linear transformation represented with respect to the standard basis on P3 by the matrix SLS = A =  2 3 13 9 6 1 6 5  . Our objective is to find a minimal spanning set for ker(L) and im(L) (with respect to the standard basis for P3). First we compute rref (A) =  1 0 −10 1 1 0 0 0  . We then see Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 65 / 73 Linear Transformations - Matrix representation Examples The column space C(A) of A is Span{A(:, 1),A(:, 2)} = Span    23 1   ,  39 6    ; the nullspace N(A) of A is Span     1−1 1    . Note that these are coordinate vectors. The corresponding vectors, written as polynomials in P3, give im(L) = Span { (2 + 3x + x2), (3 + 9x + 6x2) } , ker(L) = Span { (1− x + x2) } Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 66 / 73 Linear Transformations - Matrix representation Examples The column space C(A) of A is Span{A(:, 1),A(:, 2)} = Span    23 1   ,  39 6    ; the nullspace N(A) of A is Span     1−1 1    . Note that these are coordinate vectors. The corresponding vectors, written as polynomials in P3, give im(L) = Span { (2 + 3x + x2), (3 + 9x + 6x2) } , ker(L) = Span { (1− x + x2) } Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 66 / 73 Linear Transformations - Matrix representation Examples The column space C(A) of A is Span{A(:, 1),A(:, 2)} = Span    23 1   ,  39 6    ; the nullspace N(A) of A is Span     1−1 1    . Note that these are coordinate vectors. The corresponding vectors, written as polynomials in P3, give im(L) = Span { (2 + 3x + x2), (3 + 9x + 6x2) } , ker(L) = Span { (1− x + x2) } Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 66 / 73 Linear Transformations - Matrix representation Examples The column space C(A) of A is Span{A(:, 1),A(:, 2)} = Span    23 1   ,  39 6    ; the nullspace N(A) of A is Span     1−1 1    . Note that these are coordinate vectors. The corresponding vectors, written as polynomials in P3, give im(L) = Span { (2 + 3x + x2), (3 + 9x + 6x2) } , ker(L) = Span { (1− x + x2) } Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 66 / 73 Linear Transformations - Matrix representation Examples The column space C(A) of A is Span{A(:, 1),A(:, 2)} = Span    23 1   ,  39 6    ; the nullspace N(A) of A is Span     1−1 1    . Note that these are coordinate vectors. The corresponding vectors, written as polynomials in P3, give im(L) = Span { (2 + 3x + x2), (3 + 9x + 6x2) } , ker(L) = Span { (1− x + x2) } Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 66 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Suppose that V is an n-dimensional vector space equipped with two bases S1 = {v1, v2, . . . , vn} and S2 = {w1,w2, . . . ,wn} (as we noted above, any two bases for V must have the same number of elements). Taking L = Id : V → V the previous results yield the equation S2(v) = S2(Id ∗ v) = S2 IdS1 ∗ S1v where S2 IdS1 = [S2v1 S2v2 . . . S2vn] The matrix S2 IdS1 is referred to as a base transition matrix, and written as S2TS1 . In words, these equations tell us that in order to compute the coordinate vector S2v from S1v, we multiply S1v on the left by the n × n matrix whose i th column is the coordinate vector of vi with respect to the basis S2. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 67 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis Theorem Suppose Si , 1 ≤ i ≤ 3 are three bases for V . Then one has the following equalities 1 S3TS1 = S3TS2 ∗ S2TS1 2 Si TSi = Id 3 Si TSj = ( Sj TSi )−1 For vector spaces other than Rn (such as the function space F [a, b] we looked at earlier) where the vectors do not naturally look like column vectors, we always use the above notation when working with their coordinate representations. However, in the case of Rn, the vectors were defined as column vectors even before discussing coordinate representations. So what should we do here? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 68 / 73 Change of basis The answer is that the vectors in Rn are, by convention, identified with their coordinate representations in the standard basis e = {e1, . . . , en} for Rn. So, for example, in R3 when we wrote v =   23 −1   what we really meant was that v is the vector in R3 with coordinate representation in the standard basis given by ev =   23 −1  . Because it is very important to keep track of bases whenever determining base transition matrices and computing new coordinate representations, when doing so we will always use base-subscript notation when working with coordinate vectors, even when the vectors are in Rn and are being represented in the standard basis. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 69 / 73 Change of basis The answer is that the vectors in Rn are, by convention, identified with their coordinate representations in the standard basis e = {e1, . . . , en} for Rn. So, for example, in R3 when we wrote v =   23 −1   what we really meant was that v is the vector in R3 with coordinate representation in the standard basis given by ev =   23 −1  . Because it is very important to keep track of bases whenever determining base transition matrices and computing new coordinate representations, when doing so we will always use base-subscript notation when working with coordinate vectors, even when the vectors are in Rn and are being represented in the standard basis. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 69 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples Suppose S = {u1,u2,u3} are three vectors in R3 with eu1 =  10 2   , eu2 =  −23 1   , eu3 =   01 −1   Suppose we want to Show that the set of vectors S is a basis for R3, compute the base transition matrix STe, for v in R3 with ev =   23 −1  , compute the coordinate representation of v with repsect to the basis S. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 70 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples To perform step 1, since S has the right number of vectors to be a basis for R3, it suffices to show the vectors are linearly independent. And we know how to do this; we form the matrix A = [eu1 eu2 eu3] and show that the columns are linearly independent by showing rref (A) = Id3×3 (exercise: do this, using MATLAB or Octave). This verifies S is a basis. Next, we look at the matrix A. The columns of A are the coordinate representations of the vectors in S with respect to the standard basis e. But S is a basis. So the matrix A identifies as a base-transition matrix. We know it must be either STe or eTS . But which one? Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 71 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis Examples This is where the notation being used helps us. The coordinate vectors of the columns of A have “e” in the lower left. The rule is that this must match what appears in the notation of the transition matrix. So the only possibility is eTS = A To complete the second step, we then compute STe = (eTS)−1 Finally, we can use this to compute Sv as Sv = STe ∗ ev Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 72 / 73 Change of basis The following theorem combines base-transition in both the domain and range, together with matrix representations of linear transformations. It amounts to a “base-transition” for matrix representations of linear transformations. Theorem If S1,S ′1 are two bases for V , S2,S ′ 2 two bases for W , and L : V →W a linear transformation from V to W , then S′2 LS′1 = S′2TS2 ∗ S2LS1 ∗ S1TS′1 Note: Square matrices B,C which satisfy the equality B = A ∗ C ∗ A−1 are called similar. This is an important relation between square matrices, and plays a prominent role in the theory of eigenvalues and eigenvectors as we will see later on. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 73 / 73 Change of basis The following theorem combines base-transition in both the domain and range, together with matrix representations of linear transformations. It amounts to a “base-transition” for matrix representations of linear transformations. Theorem If S1,S ′1 are two bases for V , S2,S ′ 2 two bases for W , and L : V →W a linear transformation from V to W , then S′2 LS′1 = S′2TS2 ∗ S2LS1 ∗ S1TS′1 Note: Square matrices B,C which satisfy the equality B = A ∗ C ∗ A−1 are called similar. This is an important relation between square matrices, and plays a prominent role in the theory of eigenvalues and eigenvectors as we will see later on. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 73 / 73 Change of basis The following theorem combines base-transition in both the domain and range, together with matrix representations of linear transformations. It amounts to a “base-transition” for matrix representations of linear transformations. Theorem If S1,S ′1 are two bases for V , S2,S ′ 2 two bases for W , and L : V →W a linear transformation from V to W , then S′2 LS′1 = S′2TS2 ∗ S2LS1 ∗ S1TS′1 Note: Square matrices B,C which satisfy the equality B = A ∗ C ∗ A−1 are called similar. This is an important relation between square matrices, and plays a prominent role in the theory of eigenvalues and eigenvectors as we will see later on. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 73 / 73 Change of basis The following theorem combines base-transition in both the domain and range, together with matrix representations of linear transformations. It amounts to a “base-transition” for matrix representations of linear transformations. Theorem If S1,S ′1 are two bases for V , S2,S ′ 2 two bases for W , and L : V →W a linear transformation from V to W , then S′2 LS′1 = S′2TS2 ∗ S2LS1 ∗ S1TS′1 Note: Square matrices B,C which satisfy the equality B = A ∗ C ∗ A−1 are called similar. This is an important relation between square matrices, and plays a prominent role in the theory of eigenvalues and eigenvectors as we will see later on. Linear Algebra Module II: Vector Spaces and Linear TransformationsSummer 2021 73 / 73 Vector Spaces The space Rn Vector Spaces - Definition Subspaces Spanning Sets, Row Spaces, Column Spaces Nullspaces Range Bases and Dimension Coordinate Systems Linear Transformations Linear Transformations - Definition, properties, and terminology Linear Transformations - Matrix representation Change of basis