A First Course in Linear Algebra, 2017a

5.6. Isomorphisms 297
Proof. First suppose that T is a linear transformation and is one to one and {⃗u 1 ,···,⃗u k } is linearly independent.
It is required to show that {T (⃗u 1 ),···,T (⃗u k )} is also linearly independent. Suppose then that
Then, since T is linear,
k
∑
i=1
T
c i T (⃗u i )=⃗0
( n∑
i=1c i ⃗u i
)
=⃗0
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 span{T (⃗v 1 ),···,T (⃗v n )} = W. If
⃗w ∈ W ,thensinceT is onto 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 n
∑ c i ⃗v i =⃗v.
i=1
Hence,
⃗w = T (⃗v)=T
( n∑
i=1c i ⃗v i
)
=
n
∑
i=1
c i T (⃗v i )
It follows that span{T (⃗v 1 ),···,T (⃗v n )} = W showing that this set of vectors is a basis for W .
Next suppose that T is a linear transformation which takes a basis to a basis. This means that if
{⃗v 1 ,···,⃗v n } is a basis for V, it follows {T (⃗v 1 ),···,T (⃗v n )} is a basis for W. Thenifw ∈ W, thereexist
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 )=⃗0 then
∑ 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)=⃗0 then⃗v =⃗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 5.44: Isomorphic Subspaces
Suppose V and W are two subspaces of R n . Then the two subspaces are isomorphic if and only if
they have the same dimension. In the case that the two subspaces have the same dimension, then
for a linear map T : V → W, the following are equivalent.
1. T is one to one.
2. T is onto.
3. T is an isomorphism.