A First Course in Linear Algebra, 2017a

9.6. Linear Transformations 495
2. Let⃗v ∈ V ;then−⃗v ∈ V is the additive inverse of⃗v, so⃗v +(−⃗v)=⃗0 V . Thus
T (⃗v +(−⃗v)) = T (⃗0 V )
T (⃗v)+T (−⃗v)) = ⃗0 W
T (−⃗v) = ⃗0 W − T (⃗v)=−T (⃗v).
3. This result follows from preservation of addition and preservation of scalar multiplication. A formal
proof would be by induction on m.
Consider the following example using the above theorem.
Example 9.58: Linear Combination
Let T : P 2 → R be a linear transformation such that
Find T (4x 2 + 5x − 3).
T (x 2 + x)=−1;T (x 2 − x)=1;T (x 2 + 1)=3.
♠
Solution. We provide two solutions to this problem.
Solution 1: Suppose a(x 2 + x)+b(x 2 − x)+c(x 2 + 1)=4x 2 + 5x − 3. Then
(a + b + c)x 2 +(a − b)x + c = 4x 2 + 5x − 3.
Solving for a, b,andc results in the unique solution a = 6, b = 1, c = −3.
Thus
T (4x 2 + 5x − 3) = T ( 6(x 2 + x)+(x 2 − x) − 3(x 2 + 1) )
= 6T (x 2 + x)+T (x 2 − x) − 3T (x 2 + 1)
= 6(−1)+1 − 3(3)=−14.
Solution 2: Notice that S = {x 2 + x,x 2 − x,x 2 + 1} is a basis of P 2 , and thus x 2 , x, and 1 can each be
written as a linear combination of elements of S.
x 2 = 1 2 (x2 + x)+ 1 2 (x2 − x)
x = 1 2 (x2 + x) − 1 2 (x2 − x)
1 = (x 2 + 1) − 1 2 (x2 + x) − 1 2 (x2 − x).
Then
T (x 2 ) = T ( 1
2
(x 2 + x)+ 1 2 (x2 − x) ) = 1 2 T (x2 + x)+ 1 2 T (x2 − x)