Square root of nilpotent matrix How could I show that $\forall n \ge 2$ if $A^n=0$ and $A^{n-1} \ne 0$ then $A$ has no square root? That is there is no $B$ such that $B^2=A$. Both matrices are $n \times n$.
Thank you.

I found this post while trying to write solutions for my class's homework assignment. I had asked them precisely this question and couldn't figure it out myself. One of my students actually found a very simple solution, so I wanted to share it here:
Suppose that $N = A^2$. Since $N^n = 0$, we have $A^{2n} = 0$. Thus $A$ is nilpotent, so in fact $A^n = 0$. [This is because of the standard fact that if $P$ is any $n \times n$ nilpotent matrix, then $P^n = 0$.] Now observe that $N^{n-1} = (A^2)^{n-1} = A^{2n-2}$. But if $n \geq 2$, then $2n - 2 = n + (n - 2) \geq n$, so $N^{n-1} = 0$. This is a contradiction, so no such $A$ can exist. 
