If $A^2 = O$, is $A = O$?

Question

I think the answer is "no", but I'm trying to find the flaw in this reasoning:

$A^2 = O \implies AA = O \implies A^{-1}AA = A^{-1}O \implies A = O$

This shouldn't be true, as far as I know, so what did I do that I'm not allowed to do?

Answer 1

The problem as indicated by comments is that $A$ is not necessarily invertible. $A^{-1}$ does not necessarily exist. So from $AA = O$ you cannot simply multiply both sides by $A^{-1}$.

Answer 2

After AA = 0, You divided the 2 sides by a zero value, which is not allowed.

Something like - 2x + 1 = 0 Dividing both sides by (2x +1), 1 = 0