Answer to Question #219071 in Linear Algebra for Unknown346307

Question

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.

Answer 1

1) $T$ is an isomorphism.

We need to show that $T(\beta)$ is a basis for $W$ .

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

Let $\sum \limits_{i=1}^na_iT(\beta_i)=0$ for some $a_i$ .

Then $\sum \limits_{i=1}^na_iT(\beta_i)= \sum \limits_{i=1}^n T(a_i\beta_i)=T\left(\sum\limits_{i=1}^na_i\beta_i \right)=0$

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

Therefore, $\sum\limits_{i=1}^na_i\beta_i=0$ . Since $\beta$ is a basis for $V$ , all $a_i=0$ .

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

2) $T(\beta)$ is a basis for $W$ .

We need to show that $T$ is an isomorphism.

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

Suppose that $v\in \text{Ker} \ T$ and $v=\sum\limits_{i=1}^na_i\beta_i$ for some $a_i$ .

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

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

$T$ is surjective if and only if for all $w\in W$ there exists $v\in V$ such that $Tv=w$ .

Let $w=\sum\limits_{i=1}^na_iT(\beta_i)$ for some $a_i$ .

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

$T$ is injective and surjective. So, $T$ is an isomorphism.