Prove that $A$ cannot be invertible if $A^2=0$ 
Let $A$ be an $n\times n$ matrix for which $A^2=0$. Prove that $A$ can not be invertible.

My attempt:
Given $A^2 = 0$, this means that $A = 0$. If $A$ is invertible, there must be an $n \times n$ matrix $B$ such that $AB = I$. However, because $A = 0$, this is not possible, thus $A$ is not invertible. 
 A: If $A$ were invertible, then $BA=I$ for some $B$. Then
$$
A=IA=(BA)A=BA^2=B0=0
$$
But now $BA=B0=0$, a contradiction.
A: If $A$ is invertible, then $A^2$ is invertible too. Thus $$I=(A^2)^{-1}A^2=(A^2)^{-1}0=0,$$ a contradiction.
A: For some reason this elementary question has gotten several answers, so I'll add another. The equation $A^2=0$ implies the kernel of $A$ contains the image of $A$. By the rank-nullity theorem, it follows that the kernel of $A$ cannot be trivial.
A: Unfortunately, $A^2=0$ does not imply that $A=0$. Consider for instance $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.
Instead, if $A$ is invertible then it has an inverse $B$. What happens if you multiply both sides of $A^2=0$ by $B$ (on the left, say)?
A: $A^2=0$ $\Rightarrow$ $\det(A^2)=\det(A)^2=0$ $\Rightarrow$ $\det(A)=0$.
A: If $\pi_{i=1}^{n}A_i=0$ then atleast one $A_i$ has one eigenvalue equal to $0$, which implies atleast one of the $A_i$ is not invertible.
Alternatively, Suppose $A$ is invertible i.e. $A^{-1}$ exists. Then pre-multiply by $A^{-1}$ to both sides of $A^2=O$ gives $A=O$ (a contradiction)
