A First Course in Linear Algebra, 2017a

506 Vector Spaces
Since T is one to one, it follows that
n
∑
i=1
c i ⃗u i = 0
Now the fact that {⃗u 1 ,···,⃗u n } is linearly independent implies that each c i = 0. Hence {T (⃗u 1 ),···,T (⃗u n )}
is linearly independent.
Now suppose that T is an isomorphism and {⃗v 1 ,···,⃗v n } is a basis for V. It was just shown that
{T (⃗v 1 ),···,T (⃗v n )} is linearly independent. It remains to verify that the span of {T (⃗v 1 ),···,T (⃗v n )} is all
of W.ThisiswhereT is onto is used. If ⃗w ∈ W, there exists⃗v ∈ V such that T (⃗v)=⃗w. Since{⃗v 1 ,···,⃗v n }
is a basis, it follows that there exists scalars {c i } n i=1 such that
Hence,
⃗w = T (⃗v)=T
n
∑
i=1
c i ⃗v i =⃗v.
( n∑
i=1c i ⃗v i
)
which shows that the span of these vectors {T (⃗v 1 ),···,T (⃗v n )} is all of W showing that this set of vectors
is a basis for W.
Next suppose that T is a linear map which takes a basis to a basis. Then for {⃗v 1 ,···,⃗v n } abasis
for V ,itfollows{T (⃗v 1 ),···,T (⃗v n )} is a basis for W. Thenifw ∈ W, there exist scalars c i such that
w = ∑ n i=1 c iT (⃗v i )=T (∑ n i=1 c i⃗v i ) showing that T is onto. If T (∑ n i=1 c i⃗v i )=0then∑ n i=1 c iT (⃗v i )=⃗0 and
since the vectors {T (⃗v 1 ),···,T (⃗v n )} are linearly independent, it follows that each c i = 0. Since ∑ n i=1 c i⃗v i
is a typical vector in V , this has shown that if T (⃗v)=0then⃗v =⃗0 andsoT is also one to one. Thus T is
an isomorphism.
♠
The following theorem illustrates a very useful idea for defining an isomorphism. Basically, if you
know what it does to a basis, then you can construct the isomorphism.
Theorem 9.75: Isomorphic Vector Spaces
=
n
∑
i=1
c i T⃗v i
Suppose V and W are two vector spaces. Then the two vector spaces are isomorphic if and only if
they have the same dimension. In the case that the two vector spaces have the same dimension, then
for a linear transformation T : V → W, the following are equivalent.
1. T is one to one.
2. T is onto.
3. T is an isomorphism.
Proof. Suppose first these two vector spaces have the same dimension. Let a basis for V be {⃗v 1 ,···,⃗v n }
and let a basis for W be {⃗w 1 ,···,⃗w n }.NowdefineT as follows.
T (⃗v i )=⃗w i