More fun in 2016, Part 17

We will show that $A^{2017}=$ 0 implies $A^{2016}=$ 0 , so that there are $\boxed{0}$ solutions.

Please read this before proceeding, at least the first three lines; that's all we will need.

We are told that the polynomial $q(x)=x^{2017}$ vanishes at $A$ . Thus the minimal polynomial $p(x)$ , which divides $q(x)$ , will be of the form $p(x)=x^n$ for some $n\leq 2017$ . Since $p(x)$ also divides the characteristic polynomial of $A$ (of degree 2016), we have in fact $n\leq 2016$ , implying that $p(A)=A^{2016}=$ 0 as claimed.

More generally, we have shown: If $A$ is an $n\times n$ matrix such that $A^m=$ 0 for some positive integer $m$ (we say that $A$ is "nilpotent"), then $A^n=$ 0 .