Let $T : V → W$ be an isomorphism. Let $\{v_1,\dots, v_k\}$ be a subset of $V$. Prove that $\{v_1,\dots,v_k\}$ is a linearly independent set

Question

Let $T : V → W$ be an isomorphism. Let $\{v_1,\dots, v_k\}$ be a subset of $V$. Prove that $\{v_1,\dots, v_k\}$ is a linearly independent set if and only if $\{T(v_1),\dots,T(v_k)\}$ is a linearly independent set

I started my proof by supposing $\{v_1,\dots, v_k\}$ is a linearly independent set, but I don't see how knowing $T$ is one-to-one correspondence helps in arriving at the conclusion that $\{T(v_1),\dots,T(v_k)\}$ is linearly independent as well.

I need a little help here.

Answer 1

"$\Rightarrow$" Assume that $\{v_1,\ldots, v_n\}$ is linearly independent. Consider the equation

$$\lambda_1 T(v_1)+\cdots +\lambda_n T(v_n)=0$$

If we can prove that $\lambda_1=\cdots=\lambda_n=0$ then we are done. Since $T$ is linear then

$$T(\lambda_1 v_1+\cdots +\lambda_n v_n)=0$$

Since $T$ is one to one, then this implies that

$$\lambda_1 v_1+\cdots +\lambda_n v_n=0$$

and since $\{v_1, \ldots, v_n\}$ is linearly independent then

$$\lambda_1=\cdots=\lambda_n=0\ \Box$$

Now the reverse implication "$\Leftarrow$" actually follows from the same reasoning applied to $T^{-1}$.

Answer 2

Hint:

Any linear relation between $v_1,\dots v_k$, transformed by $T$ yields the same linear relation between $T(v_1),\dots, T(v_k)$.

The other way back, use the inverse isomorphism $T^{-1}$.

Answer 3

Let be $E = \text{span}\{v_1,\dots,v_k\}$, $F = \text{span}\{T(v_1),\dots,T(v_k)\}$. It's easy to see that $T\vert_E:E\longrightarrow F$ is an isomorphism. Let be $d = \dim E = \dim F$. Now $\{v_1,\dots,v_k\}$ is l.i. iff $k = \dim E = d$ iff $k = d = \dim F$ iff $\{T(v_1),\dots,T(v_k)\}$ is l.i.

Answer 4

You just need $T$ is injective and to apply the definitions.

Lemma. For any linear map $T\colon V\to W$, if the set $\{T(v_1),\dots,T(v_k)\}$ is linearly independent, then $\{v_1,\dots,v_k\}$ is linearly independent.

Proof. Suppose $\alpha_1v_1+\dots+\alpha_kv_k=0$. Then $$ 0=T(\alpha_1v_1+\dots+\alpha_kv_k)= \alpha_1T(v_1)+\dots+\alpha_kT(v_k) $$ so $\alpha_1=\dots=\alpha_k=0$. QED

Lemma. If $T\colon V\to W$ is an injective linear map and $\{v_1,\dots,v_k\}$ is linearly independent, then $\{T(v_1),\dots,T(v_k)\}$ is linearly independent.

Proof. Suppose $\alpha_1T(v_1)+\dots+\alpha_kT(v_k)$. Then $$ T(\alpha_1v_1+\dots+\alpha_kv_k)=0 $$ so $\alpha_1v_1+\dots+\alpha_kv_k=0$ by injectivity. Therefore $\alpha_1=\dots=\alpha_k=0$. QED