A First Course in Linear Algebra, 2017a

9.9. The Matrix of a Linear Transformation 519
2. Again remember that the order of B is important. We proceed as above. We need to find a 1 ,a 2 ,a 3
such that ⃗x = a 1 (x 2 )+a 2 (x)+a 3 (1),thatis:
Here the solution is
−x 2 − 2x + 4 = a 1 (x 2 )+a 2 (x)+a 3 (1)
a 1 = −1
a 2 = −2
a 3 = 4
Therefore the coordinate vector is
⎡
C B (⃗x)= ⎣
−1
−2
4
⎤
⎦
3. Now we need to find a 1 ,a 2 ,a 3 such that⃗x = a 1 (x + x 2 )+a 2 (x)+a 3 (4),thatis:
−x 2 − 2x + 4 = a 1 (x + x 2 )+a 2 (x)+a 3 (4)
= a 1 (x 2 )+(a 1 + a 2 )(x)+a 3 (4)
The solution is
a 1 = −1
a 2 = −1
a 3 = 1
and the coordinate vector is
⎡
C B (⃗x)= ⎣
−1
−1
1
⎤
⎦
♠
Given that the coordinate transformation C B : V → R n is an isomorphism, its inverse exists.
Theorem 9.89: Inverse of the Coordinate Isomorphism
Let V be a finite dimensional vector space with dimension n and ordered basis B = { ⃗ b 1 , ⃗ b 2 ,..., ⃗ b n }.
Then C B : V → R n is an isomorphism whose inverse,
C −1
B
: Rn → V
is given by
C −1
B
⎡
= ⎢
⎣
⎤
a 1
a 2
. ⎥
..
a n
⎦ = a 1 ⃗ b 1 + a 2
⃗ b2 + ···+ a n
⃗ bn for all
⎡
⎢
⎣
⎤
a 1
a 2
. ⎥
..
a n
⎦ ∈ Rn .