A First Course in Linear Algebra, 2017a

310 Linear Transformations
Find a basis for ker(T ) and im(T ).
Exercise 5.7.4 Let V = R 3 and let
⎧⎡
⎨
W = span ⎣
⎩
1
1
1
⎤
⎡
⎦, ⎣
−1
2
−1
⎤⎫
⎬
⎦
⎭
Extend this basis of W to a basis of V .
Exercise 5.7.5 Let T be a linear transformation given by
⎡ ⎤
x [
T ⎣ y ⎦ 1 1 1
=
1 1 1
z
What is dim(ker(T ))?
] ⎡ ⎣
x
y
z
⎤
⎦
5.8 The Matrix of a Linear Transformation II
Outcomes
A. Find the matrix of a linear transformation with respect to general bases.
We begin this section with an important lemma.
Lemma 5.55: Mapping of a Basis
Let T : R n ↦→ R n be an isomorphism. Then T maps any basis of R n to another basis for R n .
Conversely, if T : R n ↦→ R n is a linear transformation which maps a basis of R n to another basis of
R n , then it is an isomorphism.
Proof. First, suppose T : R n ↦→ R n is a linear transformation which is one to one and onto. Let {⃗v 1 ,···,⃗v n }
beabasisforR n .Wewishtoshowthat{T (⃗v 1 ),···,T (⃗v n )} is also a basis for R n .
First consider why it is linearly independent. Suppose ∑ n k=1 a kT (⃗v k )=⃗0. Then by linearity we have
T (∑ n k=1 a k⃗v k )=⃗0 and since T is one to one, it follows that ∑ n k=1 a k⃗v k =⃗0. This requires that each a k = 0
because {⃗v 1 ,···,⃗v n } is independent, and it follows that {T (⃗v 1 ),···,T (⃗v n )} is linearly independent.
Next take ⃗w ∈ R n .SinceT is onto, there exists⃗v ∈ R n such that T (⃗v)=⃗w. Since{⃗v 1 ,···,⃗v n } is a basis,
in particular it is a spanning set and there are scalars b k such that T (∑ n k=1 b k⃗v k )=T (⃗v) =⃗w. Therefore
⃗w = ∑ n k=1 b kT (⃗v k ) whichisinthespan{T (⃗v 1 ),···,T (⃗v n )}. Therefore, {T (⃗v 1 ),···,T (⃗v n )} is a basis as
claimed.
Suppose now that T : R n ↦→ R n is a linear transformation such that T (⃗v i )=⃗w i where {⃗v 1 ,···,⃗v n } and
{⃗w 1 ,···,⃗w n } are two bases for R n .