Prove that if $A^2=0$ then $A$ is not invertible Let $A$ be $n\times n$ matrix. Prove that if $A^2=\mathbf{0}$ then $A$ is not invertible.

Well I've heard that the more ways you can prove something, the merrier. :) So here's a sketch of the proof that immediately came to mind, although it may not be as snappy as some of the other good ones here:
Let's prove the contrapositive, that is if $A$ is invertible then $A^2 \neq 0$.
If $A$ is invertible then we can write it as a product of elementary matrices,
$$A = E_n...E_1I$$
Then $A^2$ can be written as
$$AA = (E_n...E_1I)(E_n...E_1I) = (E_n...E_1 E_n...E_1)I$$
which is a sequence of elementary row operations on the identity matrix. But this will never produce the zero matrix $0$. QED.
