程序代写代做代考 C Sections 9.1-9.2

Sections 9.1-9.2
Week 9
Ali Mousavidehshikh
Department of Mathematics University of Toronto
Ali Mousavidehshikh
Week 9

Outline
1 Sections 9.1-9.2
Sections 9.1-9.2
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
A set B = {b1,b2,…,bn} is called an ordered basis when it is a basis for a space in which the order of the vectors in B is taken into account.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
A set B = {b1,b2,…,bn} is called an ordered basis when it is a basis for a space in which the order of the vectors in B is taken into account. For example, {b1, b2} and {b2, b1} are not equal as ordered basis (they are the same as a basis).
Definition. If B is a basis for V and v ∈ V , then
v = 􏰀ni =1 vi bi (unique expression). The coordinate vector of
v1
v2 v is CB (v) =  . .
. vn
Week 9
Ali Mousavidehshikh

−1v2 􏰀
CB .= ni=1vibi.
. vn
Sections 9.1-9.2
Theorem: If V has dimension n and B is any ordered basis of
V , the coordinate transformation CB : V → Rn is an
isomorphism. In fact C −1 : Rn → V is given by
v1
B
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Theorem: If V has dimension n and B is any ordered basis of
V , the coordinate transformation CB : V → Rn is an
isomorphism. In fact C −1 : Rn → V is given by
v1
B
−1v2 􏰀
CB .= ni=1vibi.
. vn
If T : V → W is a linear transformation with dimV = n and
dim W = m, and B and D are ordered basis for V and W ,
respectively,thenCB :V →Rn andCD :W →Rm are
isomorphisms. In fact, the composite CD TC −1 : Rn → Rm is B
a linear transformation, so there exists a unique matrix A such that CDTC−1 = TA (where TA : Rn → Rm is given by
B
TA(x) = Ax for all x ∈ Rn).
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Hence, CD(T(v)) = TA(CB(v)) = A(CB(v)) for all v∈V.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Hence, CD(T(v)) = TA(CB(v)) = A(CB(v)) for all v ∈ V . This allows us to completely determine A.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Hence, CD(T(v)) = TA(CB(v)) = A(CB(v)) for all
v ∈ V . This allows us to completely determine A. To see this, first notice that CB (bj ) is column j of the identity matrix. So, column j of A = ACB(bj) = CD[T(bj)]. In particular,
A = [CD(T(b1)) CD(T(b2)) … CD(T(bn))].
This matrix is called the matrix of T corresponding to the ordered basis B and D and is denoted by MDB (T ). That is, we set A = MDB(T).
It follows that CD(T(v)) = MDB(T)CB(v) for all v ∈ V. Notice that MDB (T ) is an m × n matrix.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Example. Suppose T : M22 → R3 is a linear transformation 1 −1 0 0
with matrix MDB(T) = 0 1 −1 0 , where 0 0 1 −1
􏰖􏰑1 0􏰒􏰑0 1􏰒􏰑0 0􏰒􏰑0 0􏰒􏰤
B= 00,00,10,01 ,and
D = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Compute T (v) where 􏰑a b􏰒
Let V and W have ordered bases B and D and dim V = n. Then1v :V →V hasmatrixMBB(1V)=In and0:V →W has matrix MDB (0) = 0 (m × n zero matrix).
v=cd.
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
IfV→T W→S UthenST:W→U.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
IfV→T W→S UthenST:W→U.
Theorem: Let V →T W →S U be linear transformations, and let B, D, and E be finite ordered basis of V, W, and U, respectively. Then MEB(ST) = MED(S)MDB(T).
Theorem: Let T : V → W be a linear transformation, where dim V = dim W = n. The following are equivalent.
(1) T is an isomorphism.
(2) MDB (T ) is invertible for all ordered basis B and D of V and W (respectively).
(3) MDB (T ) is invertible for some pair of ordered basis B and D of V and W (respectively). When this is the case, (MDB(T))−1 = MBD(T−1).
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Example: Define T : P2 → R3 via
T(a + bx + cx2) = (a − 2b,3c − 2a,3c − 4b) for a,b,c ∈ R. Compute rank of T.
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Example: Define T : P2 → R3 via
T(a + bx + cx2) = (a − 2b,3c − 2a,3c − 4b) for a,b,c ∈ R. Compute rank of T.
Solution. Since rank T = rank(MDB (T )) for any ordered basis B and D for P2 and R3, we take the standard basis for each vector space. Then
A = MDB(T) = [CD(T(1)) CD(T(x)) CD(T(x2))]. Reducing the matrix A gives
1 −2 0 1 −2 0 1 −2 0  A=−2 0 3→0 −4 3→0 1 −3/4.
0−43 0−43 000 Hence, rank(T ) = rank(A) = 2.
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Example: Let T : V → W be a linear transformation where dim V = n and dim W = m. Choose an ordered basis
B = {b1,…,br,br+1,…,bn} of V in which {br+1,…,bn} is a basis of ker T (this set can be empty, i.e. when T is injective). Then {T (b1), . . . , T (br )} is a basis for image of T, so extend it to an ordered basis
D = {T(b1),…,T(br),fr+1,…,fm} of W. Then
MDB(T) =
[CD(T(b1)) ··· CD(T(br)) CD(T(br+1)) ··· CD(T(bn))] = 􏰑In 0􏰒
0 0 . Incidentally, this shows that rank T = r, which is the dimension of the image of T.
Definition: A linear operator is a linear transformation T : V → V (the domain and codomain are the same).
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
If T : Rn → Rn is a linear operator and E = {e1,…,en} is the standard basis for Rn, then CE (x) = x for all x ∈ Rn. We call
ME(T) = [CE(T(e1) ··· CE(T(en))] = [T(e1) ··· T(en)] the standard matrix of the operator T.
Notice that we can use all the results in the previous section by setting B = D in that section and V = W . For example, CB(T(v)) = MB(T)CB(v).
Definition: We define the change of matrix
PD←B =MDB(1V)foranyorderedbasisBandDofV.This means that PD←B = [CD(b1) ··· CD(bn)], and
CD(v) = PD←BCB(v) for all v ∈ V.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
In P2 find PD←B if B = {1,x,x2} and D={1,1−x,(1−x)2}. Thenuseittoexpress p(x)=a+bx+cx2 asapolynomialinpowerof(1−x).
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
In P2 find PD←B if B = {1,x,x2} and D={1,1−x,(1−x)2}. Thenuseittoexpress p(x)=a+bx+cx2 asapolynomialinpowerof(1−x). Solution. Notice that
1 1 1
PD←B = [CD(1) CD(x) CD(x2)] = 0 −1 −2. We have
a CB(p(x)) = b. So
c
a + b + c CD(p(x)) = PD←BCB(p(x)) =  −b − 2c . Hence,
c
p(x) = (a + b + c) − (b + 2c)(1 − x) + c(1 − x)2.
001
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Theorem. If B and D are two ordered basis of a finite dimensional vector space V and T : V → V is a linear operator, then MB(T) and MD(T) are similar matrices. More precisely, MB(T) = P−1MD(T)P, where P = PD←B is the change of matrix from B to D.
Example: Let T : R3 → R3 be given by T(a,b,c)=(2a−b,b+c,c−3a). IfD isthestandardbasis and B = {(1, 1, 0), (1, 0, 1), (0, 1, 0)}, find an invertible matrix P such that P−1MD(T)P = MB(T).
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Solution. We have MD (T ) =
[CD(2,0,−3) CD(−1,1,0) CD(0,1,1)] =  0 1 1,
4 4 −1
similarly MB(T) = −3 −2 0 , and P = PD←B =
−3 −3 2
1 1 0 [CD(1,1,0) CD(1,0,1) CD(0,1,0)] = 1 0 1. You guys
010 can verify that P−1MD(T)P = MB(T).
2 −1 0 −3 0 4
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Theorem: Let A be an n × n matrix and let E be the standard basis of Rn. From before, we know ME (TA) = A, where
TA :Rn →Rn isgivenbyTA(x)=Ax(x∈Rn asacolumn matrix).
(1) If A′ is similar to A, that is, there exists an invertible matrix P such that A′ = P−1AP and B is the ordered basis of Rn consisting of columns of P, then MB(TA) = A′.
(2) If B is any ordered basis of Rn, let P be the matrix whose columns are the vectors in B (notice this forces P to be invertible). Then MB(TA) = P−1AP.
Week 9
Ali Mousavidehshikh

Sections 9.1-9.2
Example: (a) If T : V → V is an operator where V is finite dimensional, show that TST = T for some invertible operator S : V → V . (b) If A is an n × n matrix, show that AUA = A for some invertible matrix U.
Proof. (a)LetB={b1,…,br,br+1,…,bn}beanordered basis for V chosen so that kerT = span{br+1,…,bn} (the set is automatically a basis for ker T since it is a subset of a basis). Then {T (b1), . . . , T (br )} is a linearly independent set (why?), so can be extended to a basis for V , say {T(b1),…,T(br),fr+1,…,fn}. Define S : V → V via S(T(bi))=bi for1≤i≤randS(fi)=bi forr+1≤i≤n. Then S is an isomorphism (it sends basis to basis). Moreover, TST = T (look at each sides actions on the basis bi ).
Ali Mousavidehshikh
Week 9

Sections 9.1-9.2
Definition: If T : V → V is a linear operator, we define
det T = det MB (T ), where B is any ordered basis for V (they all give the same answer, since if D is any other ordered basis of V, the matrices MB(T) and MD(T) are similar matrices). We define the trace tr(T ) = tr(MB (T )). It follows that det(ST ) = det S det T . The characteristic polynomial of T , CT(x) = CA(x) where A = MB(T) for any ordered basis B of V.
Example: Compute the determinant, trace, and characteristic polynomial of the operator T : P2 → P2 given by
T(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2.
Ali Mousavidehshikh
Week 9

Related Posts