程序代写代做代考 AI Week 2: Sections 6.3 and 6.4

Week 2: Sections 6.3 and 6.4
Week 2
Ali Mousavidehshikh
Department of Mathematics University of Toronto
Ali Mousavidehshikh
Week 2

Outline
Week 2: Sections 6.3 and 6.4
1 Week 2: Sections 6.3 and 6.4
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
A set of vectors that is not linearly independent is said to be
linearly dependent (or simply dependent).
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
A set of vectors that is not linearly independent is said to be
linearly dependent (or simply dependent).
(Example) The set {sinx,cosx} is independent in the vector space F[0, 2π].
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
A set of vectors that is not linearly independent is said to be
linearly dependent (or simply dependent).
(Example) The set {sinx,cosx} is independent in the vector space F[0, 2π].
Solution. Suppose that a1 sin x + a2 cos x = 0.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
A set of vectors that is not linearly independent is said to be
linearly dependent (or simply dependent).
(Example) The set {sinx,cosx} is independent in the vector space F[0, 2π].
Solution. Suppose that a1 sin x + a2 cos x = 0.This must hold for all values of x ∈ [0, 2π].
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
A set of vectors {v1,v2,…,vn} in a vector space V is called linearly independent (or simply independent) if its satisfies the following condition:
n
If 􏰁 ai vi = 0, then ai = 0 for i = 1, 2, . . . , n.
i=1
A set of vectors that is not linearly independent is said to be
linearly dependent (or simply dependent).
(Example) The set {sinx,cosx} is independent in the vector space F[0, 2π].
Solution. Suppose that a1 sin x + a2 cos x = 0.This must hold forallvaluesofx∈[0,2π]. Takingx=0givess2=0and
takingx=π givess1=0. 2
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
Hard example. Suppose p1, p2, . . . , pn are polynomials, degpi =di withd1 >d2 >…>dn. Provethat
{p1, p2, . . . , pn} is a linearly independent set.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Hard example. Suppose p1, p2, . . . , pn are polynomials, degpi =di withd1 >d2 >…>dn. Provethat
{p1, p2, . . . , pn} is a linearly independent set. Proof .
Suppose􏰀ni=1tipi =0.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Hard example. Suppose p1, p2, . . . , pn are polynomials, degpi =di withd1 >d2 >…>dn. Provethat
{p1, p2, . . . , pn} is a linearly independent set. Proof .
Suppose􏰀ni=1tipi =0.
Let axd1 be the term in p1 of the highest degree (notice a ̸= 0).
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Hard example. Suppose p1, p2, . . . , pn are polynomials, degpi =di withd1 >d2 >…>dn. Provethat
{p1, p2, . . . , pn} is a linearly independent set. Proof .
Suppose􏰀ni=1tipi =0.
Let axd1 be the term in p1 of the highest degree (notice a ̸= 0). Then t1axd1 is the only term of degree d1 in the above linear combination.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Hard example. Suppose p1, p2, . . . , pn are polynomials, degpi =di withd1 >d2 >…>dn. Provethat
{p1, p2, . . . , pn} is a linearly independent set. Proof .
Suppose􏰀ni=1tipi =0.
Let axd1 be the term in p1 of the highest degree (notice a ̸= 0). Then t1axd1 is the only term of degree d1 in the above linear combination.
Hence, t1 = 0. Repeat the process to show that ti = 0 for all
i = 1,2,…,n.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
SupposeAisann×nmatrixs.t. Ak =0butAk−1 ̸=0. Prove that {I,A,…Ak−1} is a linearly independent set.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
SupposeAisann×nmatrixs.t. Ak =0butAk−1 ̸=0. Prove that {I,A,…Ak−1} is a linearly independent set. Proof .
Supposea0I+a1A+…ak−1Ak−1 =0.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
SupposeAisann×nmatrixs.t. Ak =0butAk−1 ̸=0. Prove that {I,A,…Ak−1} is a linearly independent set. Proof .
Supposea0I+a1A+…ak−1Ak−1 =0.
Multiply both sides by Ak−1, and conclude that a0 = 0.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
SupposeAisann×nmatrixs.t. Ak =0butAk−1 ̸=0. Prove that {I,A,…Ak−1} is a linearly independent set. Proof .
Supposea0I+a1A+…ak−1Ak−1 =0.
Multiply both sides by Ak−1, and conclude that a0 = 0. Now multiply by Ak−2 to conclude that a1 = 0.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
SupposeAisann×nmatrixs.t. Ak =0butAk−1 ̸=0. Prove that {I,A,…Ak−1} is a linearly independent set. Proof .
Supposea0I+a1A+…ak−1Ak−1 =0.
Multiply both sides by Ak−1, and conclude that a0 = 0. Now multiply by Ak−2 to conclude that a1 = 0. Continue this process to show that ai = 0 for all
i = 0, 1, . . . k − 1.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Theorem. Suppose V is spanned by n vectors. If any set of m vectors in V is linearly independent, then m ≤ n.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Theorem. Suppose V is spanned by n vectors. If any set of m vectors in V is linearly independent, then m ≤ n.
Definition. A set {v1, v2, . . . , vn} is called a basis for a vector space V if it satisfies the following two conditions:
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Theorem. Suppose V is spanned by n vectors. If any set of m vectors in V is linearly independent, then m ≤ n.
Definition. A set {v1, v2, . . . , vn} is called a basis for a vector space V if it satisfies the following two conditions:
(1) V = span{v1,v2,…,vn},
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Theorem. Suppose V is spanned by n vectors. If any set of m vectors in V is linearly independent, then m ≤ n.
Definition. A set {v1, v2, . . . , vn} is called a basis for a vector space V if it satisfies the following two conditions:
(1) V = span{v1,v2,…,vn},
(2) The set {v1, v2, . . . , vn} is independent.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
(Example) If v ̸= 0, then {v} is a linearly independent set. No independent set of vectors in V can contain the zero vector.
Theorem. Let {v1, v2, . . . , vn} be a linear independent set. If 􏰀ni=1 aivi = 􏰀ni=1 bivi, then ai = bi for all i = 1,2,…n.
Theorem. Suppose V is spanned by n vectors. If any set of m vectors in V is linearly independent, then m ≤ n.
Definition. A set {v1, v2, . . . , vn} is called a basis for a vector space V if it satisfies the following two conditions:
(1) V = span{v1,v2,…,vn},
(2) The set {v1, v2, . . . , vn} is independent.
Theorem. If {e1,e2,…,en} and {f1,f2,…,fm} is a basis of V , then n = m.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn,
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn, dimRn =n,
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn, dimRn =n, dimPn =n+1,
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn, dimRn =n, dimPn =n+1, dimP = ∞,
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn,
dimRn =n,
dimPn =n+1,
dimP = ∞,
dim span{v } = 1, provided v is a non-zero vector.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
WesaydimV =nifithasabasiswithnelementsinit(we call such spaces finite dimensional).
We say dim V = ∞ if it has basis with infinitely many elements (we call such spaces infinite dimensional).
We define the dimension of the zero vector space to be zero.
Theorem.
dimMmn =mn,
dimRn =n,
dimPn =n+1,
dimP = ∞,
dim span{v } = 1, provided v is a non-zero vector.
What do you think is the dimension of all 2 × 2 symmetric matrices?
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Theorem. Let V be a finite dimensional vector space spanned by m vectors.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Theorem. Let V be a finite dimensional vector space spanned by m vectors. Then
(a) V has a finite basis and dimV ≤ m.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Theorem. Let V be a finite dimensional vector space spanned by m vectors. Then
(a) V has a finite basis and dimV ≤ m.
(b) Every independent set of vectors in V can be enlarged to a
basis of V by adding some vectors from any fixed basis of V .
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Theorem. Let V be a finite dimensional vector space spanned by m vectors. Then
(a) V has a finite basis and dimV ≤ m.
(b) Every independent set of vectors in V can be enlarged to a
basis of V by adding some vectors from any fixed basis of V .
(c) If U is a subspace of V , then U is finite dimensional, and
dim U ≤ dim V . Moreover, every basis of U is part of a basis of V.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. Let V be a finite dimensional vector space. If U is any subspace of V , then any independent subset of U can be enlarged to a finite basis of U.
Theorem. Let V be a finite dimensional vector space spanned by m vectors. Then
(a) V has a finite basis and dimV ≤ m.
(b) Every independent set of vectors in V can be enlarged to a
basis of V by adding some vectors from any fixed basis of V .
(c) If U is a subspace of V , then U is finite dimensional, and
dim U ≤ dim V . Moreover, every basis of U is part of a basis of V.
The set {(1, 0, 0), (1, 1, 0)} is independent set. Enlarge it to a basis for R3.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W. Proof .
Noticethat(x−a),(x−a)2,…,(x−a)n ∈W.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W. Proof .
Noticethat(x−a),(x−a)2,…,(x−a)n ∈W.
Moreover, they are independent (they have different degrees).
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W. Proof .
Noticethat(x−a),(x−a)2,…,(x−a)n ∈W.
Moreover, they are independent (they have different degrees). Let U = span{(x − a), (x − a)2, . . . , (x − a)n}.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W. Proof .
Noticethat(x−a),(x−a)2,…,(x−a)n ∈W.
Moreover, they are independent (they have different degrees). Let U = span{(x − a), (x − a)2, . . . , (x − a)n}.
ThenU⊆W ⊆Pn,withdimU=nanddimPn =n+1.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Theorem. Let U and W be subspace of a finite dimensional vector space V .
(a) IfU⊆W,thendimU⊆dimW.
(b) IfU⊆W anddimU=dimW,thenU=W.
LetW ={P(x)∈Pn :p(a)=0}. Showthat
{(x −a),(x −a)2,…,(x −a)n} is a basis for W. Proof .
Noticethat(x−a),(x−a)2,…,(x−a)n ∈W.
Moreover, they are independent (they have different degrees). Let U = span{(x − a), (x − a)2, . . . , (x − a)n}.
ThenU⊆W ⊆Pn,withdimU=nanddimPn =n+1. Hence,U=W orW =Pn+1. SinceW ̸=Pn (why?),U=W.
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
Lemma. A set D = {v1,…,vn} of vectors in a vector space V is dependent if and only if some vector in D is a linear combination of the others.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. A set D = {v1,…,vn} of vectors in a vector space V is dependent if and only if some vector in D is a linear combination of the others.
Lemma. Let V be a finite dimensional vector space. Any spanning set for V can be cut down to a basis of V.
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. A set D = {v1,…,vn} of vectors in a vector space V is dependent if and only if some vector in D is a linear combination of the others.
Lemma. Let V be a finite dimensional vector space. Any spanning set for V can be cut down to a basis of V.
Theorem. Let V be a vector space with dim V = n, and suppose S is a subset of V with exactly n vectors. Then S is linearly independent if and only if S spans V .
Week 2
Ali Mousavidehshikh

Week 2: Sections 6.3 and 6.4
Lemma. A set D = {v1,…,vn} of vectors in a vector space V is dependent if and only if some vector in D is a linear combination of the others.
Lemma. Let V be a finite dimensional vector space. Any spanning set for V can be cut down to a basis of V.
Theorem. Let V be a vector space with dim V = n, and suppose S is a subset of V with exactly n vectors. Then S is linearly independent if and only if S spans V .
Theorem. Suppose U and W are finite dimensional subspaces of a vector space V . Then U + W is a finite dimensional subspace of V
Ali Mousavidehshikh
Week 2

Week 2: Sections 6.3 and 6.4
Lemma. A set D = {v1,…,vn} of vectors in a vector space V is dependent if and only if some vector in D is a linear combination of the others.
Lemma. Let V be a finite dimensional vector space. Any spanning set for V can be cut down to a basis of V.
Theorem. Let V be a vector space with dim V = n, and suppose S is a subset of V with exactly n vectors. Then S is linearly independent if and only if S spans V .
Theorem. Suppose U and W are finite dimensional subspaces of a vector space V . Then U + W is a finite dimensional subspace of V and dim(U+W)=dimU+dimW −dim(U∩W).
Ali Mousavidehshikh
Week 2