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 [![enter image description here](https://i.stack.imgur.com/ldJfi.png)](https://i.stack.imgur.com/ldJfi.png) 

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?