If $Ax=0$ and $A$ is an invertible matrix, then $x=0$ is the unique solution?
I think it is a very basic question but I forget some knowledge in linear algebra. For an invertible $A$, the inverse $A^{-1}$ is well-defined, thus we can left-multiply $A^{-1}$ to each side of the equation to get $A^{-1}Ax=A^{-1}0$, which leads to $Ix=0$ and thus $x=0$. I think this calculation says $x=0$ must be a solution but dose not guarantee the uniqueness. It is not a proof.
Best Answer
I think you're thinking wrongly about your proof, but that doesn't show up too much in the proof itself.
The proof is fine. Except the two last sentences of course: it shows not that $x=0$ is a solution, but that no other solutions exists. And the last sentece is also wrong as it is a proof.
To be a bit elaborate we have that if $u$ is a solution, that is $Au=0$. Left multiplication by $A^{-1}$ implies that $Iu = A^{-1}Au = A^{-1}0 = 0$. And this in turn implies(*) that $u=0$.
Note that this hasn't shown that $u=0$ is a solution, but rather that if $u$ is a solution it must be $0$. However hopefully you already know that's an solution (and you also have that each step in the above proof is an equivalence, but that's not at obvious as they are implications).
(*) That $Iu =0$ implies that $u=0$ is for example shown by RAA by in addition to $Iu=0$ assuming that $u\ne 0$ nand then using the property of the identity that is $Iu = u = \ne0$ which contradicts the assumtion $Iu=0$.