Prove that resultant vectors of a linear transformation forms a basis if it is invertible. $\{v_1......v_n\}$ 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?
 A: 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.
A: 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 
$$...T(\lambda_1v_1 + ...+\lambda_nv_n) = 0\iff ST(\lambda_1v_1 + ...+\lambda_nv_n) = 0\iff \lambda_1v_1 + ...+\lambda_nv_n = 0 \iff \lambda_1 = ....=\lambda_n = 0$$
A: 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:


*

*There exists $S$ such that $ST = \operatorname{Id}_V$.

*$T$ is monomorphism.

*For every linearly independent set $\{v_1,\ldots , v_n\}$, $\{Tv_1,\ldots,Tv_n\}$ is linearly independent.

*$T$ 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.
