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.

So $B^{2n} = 0$, but $B^{2n-2} \ne 0$.  So the Jordon Canonical Form of $B$ must contain a Jordon block whose size is at least $(2n-1)\times(2n-1)$.  And since $n \ge 2$, there isn't enough room.
