A First Course in Linear Algebra, 2017a

316 Linear Transformations
Hence
⎡
A = ⎣
1 1 0
−1 2 1
1 −1 1
⎤⎡
⎦⎣
1 1 −1
0 1 1
1 1 0
⎤
⎦
−1
⎡
= ⎣
0 0 1
2 3 −3
−3 −2 4
Of course this is a very different matrix than the matrix of the linear transformation with respect to the non
standard basis.
♠
Exercises
{[
Exercise 5.8.1 Let B =
C B (⃗x).
⎧⎡
⎨
Exercise 5.8.2 Let B = ⎣
⎩
in R 2 .FindC B (⃗x).
2
−1
1
−1
2
] [
3
,
2
⎤
⎡
⎦, ⎣
]}
[
be a basis of R 2 and let ⃗x =
2
1
2
⎤
⎡
⎦, ⎣
−1
0
2
5
−7
⎤⎫
⎡
⎬
⎦
⎭ be a basis of R3 and let ⃗x = ⎣
([ ]) a
Exercise 5.8.3 Let T : R 2 ↦→ R 2 be a linear transformation defined by T =
b
Consider the two bases
{[ 1
B 1 = {⃗v 1 ,⃗v 2 } =
0
] [ −1
,
1
]}
⎤
⎦
]
be a vector in R 2 .Find
5
−1
4
⎤
[ a + b
a − b
⎦ be a vector
]
.
and
{[
1
B 2 =
1
] [
,
1
−1
]}
Find the matrix M B2 ,B 1
of T with respect to the bases B 1 and B 2 .
5.9 The General Solution of a Linear System
Outcomes
A. Use linear transformations to determine the particular solution and general solution to a system
of equations.
B. Find the kernel of a linear transformation.
Recall the definition of a linear transformation discussed above. T is a linear transformation if
whenever⃗x,⃗y are vectors and k, p are scalars,
T (k⃗x + p⃗y)=kT (⃗x)+pT (⃗y)
Thus linear transformations distribute across addition and pass scalars to the outside.