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$\{v_1......v_n\}$ is a basis of V over F.
$T$ is a linear transformation.
$\{w_1......w_n\}$ is a set of vectors such that $T(v_i) = w_i$

Prove that if T is invertible, $\{w_1......w_n\}$ forms a basis.

Here is my attempt.
Proving that $\{w_1......w_n\}$ are linear independent. $$ \lambda_1w_1 + ...+\lambda_nw_n = 0 \\ T(\lambda_1v_1 + ...+\lambda_nv_n) = 0 \\ $$ Since $T$ is invertible, $T(v)=0$ has only one solution, $v=0$ $$ \lambda_1v_1 + ...+\lambda_nv_n = 0 \\ \lambda_1 = ....=\lambda_n = 0 $$

Is this proof correct? Could it be more rigorous?

If instead of $T$ being invertible, we are just given that, there exists another linear transformation $S$, such that $ST=I$, is it still possible to prove the same result?

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You didn't specify explicitly, but I'm assuming that $T\colon V\to V$. In this case you are right, here are some things that are equivalent when $V$ is finite-dimensional:

  1. There exists $S$ such that $ST = \operatorname{Id}_V$.
  2. $T$ is monomorphism.
  3. For every linearly independent set $\{v_1,\ldots , v_n\}$, $\{Tv_1,\ldots,Tv_n\}$ is linearly independent.
  4. $T$ is isomorphism.

However, if $V$ is not finite dimensional, or maybe more relevant for you at the moment, when $T\colon V\to W$, the numbers 1.-3. are still equivalent, but 4. is now stronger. This is because $T$ can fail to be epimorphism.

Thus, in the case $T\colon V\to W$, your proof is incomplete, you still need to argue that $\{Tv_1,\ldots,Tv_n\}$ spans $W$. This is where $T$ being invertible is crucial, compared to just being monomorphism.

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  • $\begingroup$ Is there any condition for n for 3 to be equivalent to others? Perhaps n should be equal to dimension of V. $\endgroup$ – sourav goyal Apr 6 '18 at 14:29
  • $\begingroup$ @sourav goyal, notice the wording "for every linearly independent set" which explicitly allows for different $n$'s. But it should be enough to fix $n$ to be the dimension if the space is finite-dimensional. If it's not, then it would be nonsense. $\endgroup$ – Ennar Apr 6 '18 at 14:34
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Yes, the proof is correct. And it is still true if you assume that there is a linear map $S$ such that $S\circ T=\operatorname{Id}$, since it follows from this that $T$ is injective and that was the only proerty of $T$ (apart from being linear) that you used.

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  • $\begingroup$ How does it follow from $S∘T=I $, that $T$ is injective ? $\endgroup$ – sourav goyal Mar 24 '18 at 14:13
  • $\begingroup$ @souravgoyal Because$$T(v)=0\implies S\bigl(T(v)\bigr)=0\iff\operatorname{Id}(v)=0\iff v=0.$$ $\endgroup$ – José Carlos Santos Mar 24 '18 at 14:35
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Yes your proof is correct even if you could make it a bit more precise clarifying better each step as for example

$$\lambda_1w_1 + ...+\lambda_nw_n = \lambda_1T(v_1) + ...+\lambda_nT(v_n)= T(\lambda_1v_1) + ...+T(\lambda_nv_n) = T(\lambda_1v_1 + ...+\lambda_nv_n) = 0\iff \lambda_1v_1 + ...+\lambda_nv_n = 0 \iff \lambda_1 = ....=\lambda_n = 0$$

Yes we can proof the same result if we are just given that there exists another linear transformation $S$, such that $ST=I$ indeed

$$...T(\lambda_1v_1 + ...+\lambda_nv_n) = 0\iff ST(\lambda_1v_1 + ...+\lambda_nv_n) = 0\iff \lambda_1v_1 + ...+\lambda_nv_n = 0 \iff \lambda_1 = ....=\lambda_n = 0$$

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