A First Course in Linear Algebra, 2017a

522 Vector Spaces
Therefore C B2 [T (x 3 )] =
⎡
⎢
⎣
1
0
1
1
⎤
⎥
⎦ .
You can verify that the following are true.
⎡ ⎤
1
C B2 [T(x 2 )] = ⎢ 1
⎣ 1
0
a 4 = 1
⎥
⎦ ,C B 2
[T(x)] =
⎡
⎢
⎣
0
−1
−1
0
Using these vectors as the columns of M B2 B 1
(T ) we have
⎡
M B2 B 1
(T )=
⎢
⎣
⎤
1 1 0 0
0 1 −1 0
1 1 −1 −1
1 0 0 1
⎥
⎦ ,C B 2
[T (1)] =
⎤
⎥
⎦
⎡
⎢
⎣
0
0
−1
1
⎤
⎥
⎦
The next example demonstrates that this method can be used to solve different types of problems. We
will examine the above example and see if we can work backwards to determine the action of T from the
matrix M B2 B 1
(T ).
Example 9.92: Finding the Action of a Linear Transformation
Let T : P 3 ↦→ R 4 be an isomorphism with
M B2 B 1
(T )=
where B 1 = { x 3 ,x 2 ,x,1 } is an ordered basis of P 3 and
⎧⎡
⎪⎨
B 2 = ⎢
⎣
⎪⎩
1
0
1
0
⎤
⎡
⎢
⎣
⎡
⎥
⎦ , ⎢
⎣
0
1
0
0
1 1 0 0
0 1 −1 0
1 1 −1 −1
1 0 0 1
⎤
⎡
⎥
⎦ , ⎢
⎣
0
0
−1
0
⎤
⎡
⎥
⎦ , ⎢
is an ordered basis of R 4 .Ifp(x)=ax 3 + bx 2 + cx + d, findT (p(x)).
⎣
⎤
⎥
⎦ ,
0
0
0
1
⎤⎫
⎪⎬
⎥
⎦
⎪⎭
♠
Solution. Recall that C B2 [T (p(x))] = M B2 B 1
(T)C B1 (p(x)). Thenwehave
C B2 [T(p(x))] = M B2 B 1
(T )C B1 (p(x))