Page 251 - Matrix Analysis & Applied Linear Algebra

246 Chapter 4 Vector Spaces

Connections with Matrix Algebra

• If T, L ∈L(U, V), and if B and B are bases for U and V, then
[αT] BB = α[T] BB for scalars α, (4.7.7)
[T + L] BB =[T] BB +[L] BB . (4.7.8)

• If T ∈L(U, V) and L ∈L(V, W), and if B, B , and B are bases
for U, V, and W, respectively, then LT ∈L(U, W), and
[LT] BB =[L] B B [T] BB . (4.7.9)

• If T ∈L(U, U) is invertible in the sense that TT −1 = T −1 T = I
for some T −1 ∈L(U, U), then for every basis B of U,
[T −1 ] B =[T] −1 . (4.7.10)
B

Proof. The first three properties (4.7.7)–(4.7.9) follow directly from (4.7.6). For
example, to prove (4.7.9), let u be any vector in U, and write
" # " # " #
[LT] BB [u] B = LT(u) B = L T(u) B =[L] B B T(u) B =[L] B B [T] BB [u] B .
This is true for all u ∈U, so [LT] BB =[L] B B [T] BB (see Exercise 3.5.5).
Proving (4.7.7) and (4.7.8) is similar—details are omitted. To prove (4.7.10),
note that if dim U = n, then [I] B = I n for all bases B, so property (4.7.9)
implies I n =[I] B =[TT −1 ] B =[T] B [T −1 ] B , and thus [T −1 ] B =[T] −1 .
B
Example 4.7.6
Problem: Form the composition C = LT of the two linear transformations
3
2
2
2
T : → and L : → defined by
T(x, y, z)=(x + y, y − z) and L(u, v)=(2u − v, u),
2
and then verify (4.7.9) and (4.7.10) using the standard bases S 2 and S 3 for
3
and , respectively.
3
2
Solution: The composition C : → is the linear transformation

C(x, y, z)= L T(x, y, z) = L(x + y, y − z)=(2x + y + z, x + y).
The coordinate matrix representations of C, L, and T are

211 2 −1 11 0
= , = , and = .
[C] S 3 S 2 [L] S 2 [T] S 3 S 2
110 1 0 01 −1