# If $X$ is an $n$ by $n$ matrix with only $0$ as eigenvalue, why does $X^k≠0$ implies $k<n$?

The exercise is to prove how many matrices $X$ are such that $X^m=A$, knowing that $A$ have all $3$ eingenvalues $0$ and that $A^2≠0$. For $m=1$ we have $X=A$, but for $m>1$, if what I want to prove is true, there are no solutions, but I couldn't to prove this result.

• The characteristic polynomial has degree $n$, so it is $x^n$ Aug 22 '18 at 0:51
• It follows from the Cayley-Hamilton theorem Aug 22 '18 at 1:08
• See this answer Aug 22 '18 at 1:10