Suppose that $A\vec{x} = \vec{0}$ is a linear system. Also columns of $A$ are linearly independent. Prove that $\vec{x} = \vec{0}$ is the solution and it's unique.
My answer: If $A$ is a square matrix, then it has a inverse because of independent columns. So $A\vec{x} = \vec{0}$ implies that $A^{-1}A\vec{x} = A^{-1}\vec{0} = \vec{0}$ and then $\vec{x} = \vec{0}$.
If $A$ is not a square matrix, it has a left inverse because of independent columns. So $A\vec{x} = \vec{0}$ implies that $A^{-1}A\vec{x} = A^{-1}\vec{0} = \vec{0}$ and then $\vec{x} = \vec{0}$. Here $A^{-1}$ denotes the left inverse.
My professor said that second part of my argument isn't complete and it only shows that $\vec{x} = \vec{0}$ is a solution and uniqueness hasn't been proved yet. I don't get that. Is that true?