Question #219071

Let V and W be n−dimensional vector spaces, and let T : V → W be a linear transformation.

Suppose β is a basis for V . Prove that T is an isomorphism if and only if T(β) is a basis for W.



1
Expert's answer
2021-07-21T07:26:24-0400

1) TT is an isomorphism.

We need to show that T(β)T(\beta) is a basis for WW .

The set T(β)={T(β1),,T(βn)}T(\beta)=\{T(\beta_1), …, T(\beta_n)\} consists of nn vectors. It suffices to show that T(β1),,T(βn)T(\beta_1), …, T(\beta_n) are linearly independent.

Let i=1naiT(βi)=0\sum \limits_{i=1}^na_iT(\beta_i)=0 for some aia_i .

Then i=1naiT(βi)=i=1nT(aiβi)=T(i=1naiβi)=0\sum \limits_{i=1}^na_iT(\beta_i)= \sum \limits_{i=1}^n T(a_i\beta_i)=T\left(\sum\limits_{i=1}^na_i\beta_i \right)=0

Since TT is an isomorphism, it follows that Tv=0Tv=0 only if v=0v=0 .

Therefore, i=1naiβi=0\sum\limits_{i=1}^na_i\beta_i=0 . Since β\beta is a basis for VV , all ai=0a_i=0 .

Hence, T(β)T(\beta) is linearly independent.


2) T(β)T(\beta) is a basis for WW .

We need to show that TT is an isomorphism.


TT is injective if and only if Ker T=0\text{Ker}\ T=0 .

Suppose that vKer Tv\in \text{Ker} \ T and v=i=1naiβiv=\sum\limits_{i=1}^na_i\beta_i for some aia_i .

Tv=T(i=1naiβi)=i=1nT(aiβi)=i=1naiT(βi)=0.Tv=T\left(\sum\limits_{i=1}^na_i\beta_i \right)=\sum\limits_{i=1}^nT(a_i\beta_i)=\sum\limits_{i=1}^na_iT(\beta_i)=0.

Since T(β)T(\beta) is a basis for WW , it follows that all ai=0a_i=0 . Then we have that v=0v=0 and Ker T=0\text{Ker}\ T=0 .


TT is surjective if and only if for all wWw\in W there exists vVv\in V such that Tv=wTv=w .

Let w=i=1naiT(βi)w=\sum\limits_{i=1}^na_iT(\beta_i) for some aia_i .

Then w=i=1naiT(βi)=i=1nT(aiβi)=T(i=1naiβi)w=\sum\limits_{i=1}^na_iT(\beta_i)=\sum\limits_{i=1}^nT(a_i\beta_i)=T\left( \sum\limits_{i=1}^na_i\beta_i \right) , where i=1naiβiV.\sum\limits_{i=1}^na_i\beta_i\in V.


TT is injective and surjective. So, TT is an isomorphism.


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