Page 259 - Matrix Analysis & Applied Linear Algebra

254 Chapter 4 Vector Spaces

Proof. Let B = {x 1 , x 2 ,..., x n } and B = {y 1 , y 2 ,..., y n } , and observe that

for each j, (4.7.6) can be used to write

* + * +
A(x j ) =[A] B [x j ] B =[A] B P ∗j = [A] B P .
B ∗j
Now use the change of coordinates rule (4.8.4) together with the fact that
" # " #
A(x j ) = [A] B (see (4.7.4)) to write
B ∗j
* + * + * + * +
A(x j ) = P A(x j ) = P [A] B = P[A] B .
B B ∗j ∗j

" # " #
Consequently, [A] B P = P[A] B for each j, so [A] B P = P[A] B . Since
∗j ∗j
the change of basis matrix P is nonsingular, it follows that [A] B = P −1 [A] B P,
and (4.8.5) is proven. Setting Q = P −1 in (4.8.5) yields [A] B = Q −1 [A] B Q.
The matrix Q = P −1 is the change of basis matrix from B to B because if T

is the change of basis operator from B to B (i.e., T(y i )= x i ), then T −1 is

the change of basis operator from B to B (i.e., T −1 (x i )= y i ), and according

to (4.8.3), the change of basis matrix from B to B is

[I] B B = [y 1 ] B [y 2 ] B ··· [y n ] B =[T −1 ] B =[T] −1 = P −1 = Q.

B
Example 4.8.2
2
Problem: Consider the linear operator A(x, y)=(y, −2x +3y)on along
with the two bases

1 0 1 1
S = , and S = , .

0 1 1 2
First compute the coordinate matrix [A] S as well as the change of basis matrix
Q from S to S, and then use these two matrices to determine [A] S .

Solution: The matrix of A relative to S is obtained by computing

A(e 1 )=A(1, 0) = (0, −2) = (0)e 1 +(−2)e 2 ,
A(e 2 )=A(0, 1) = (1, 3) = (1)e 1 +(3)e 2 ,

0 1
so that [A] S = [A(e 1 )] S [A(e 2 )] S = . According to (4.8.6), the
−2 3

change of basis matrix from S to S is

11

Q = [y 1 ] S [y 2 ] S = ,

12