Any Flaw?) If columns of a square matrix are linearly independent, A is invertible.

Question

My strategy to prove this part is I assume columns of Matrix are linearly independent and A is not invertible,

$AX=V$ has non unique solutions, say $X_{1}$ and $X_{2}$.

Then, There are two solutions $X_{1} \neq X_{2}$ such that $AX_{1}=V$ and $AX_{2}=V$. Then if follows that $A(X_{1}-X_{2})=0$.

From which we desire a series of linear combinations of

It implies that each $\alpha_{i}$ is zero since vectors of A are linearly independent and thus $X_{1}=X_{2}$

Can I conclude from this result that A must have unique solution $X$ satisfying $AX=B$ and therefore $A$ is invertible?

Answer 1

If columns are lineraly independent $Ax=y$ is injective and surjective thus the transformation is invertible, that is exists $B=A^{-1}$ such that $x=By$.

Answer 2

You need to emphasize that your matrix is a square matrix.

Otherwise speaking of inverse in problematic.

For square matrices your argument is valid, you are showing that there are no zero eigenvalues, thus the matrix is invertible.