# A problem on matrix theory .

If $$A$$ is an $$n×n$$ square matrix over $$\mathbb C$$ and $$S(A)=\{X\in M_{n,p}(\mathbb C) : AX=0 \}$$, then there exists $$r\in\mathbb N$$ such that $$S(A^r)=S(A^{r+1})=S(A^{r+2})=\cdots$$.

Actually, it is obvious that $$S(A)$$ is a subspace of the whole $$n×p$$ matrix space over $$\mathbb C$$, and its dimension is $$2np$$. I think if an infinite chain of distinct $$n\times p$$ matrices like $$X_1, AX_2, A^2X_3,\cdots,A^kX_{k+1},\cdots$$ all in $$S(A)$$ appears where $$A^{k+1}X_{k+1} =0$$, then there will be infinitely many linearly independent matrices defending the space's finite dimension. But I can't further proceed.

A usual dimension argument will do the trick. Observe that $$A^kX=0$$ implies that $$A^{k+1}X=0$$. Therefore $$S(A^k)\subseteq S(A^{k+1})$$ for every positive integer $$n$$, i.e. $$\{ S(A^k)\}_{k\in\mathbb N}$$ is an ascending chain of subspaces. Since $$M_{n,p}(\mathbb C)$$ is finite-dimensional, the dimensions of those subspaces in the chain cannot grow indefinitely. So, the dimensions must eventually become constant at some point, i.e. there exists an $$r$$ such that $$\dim S(A^r)=\dim S(A^k)$$ for all $$k\ge r$$. However, as $$S(A^r)\subseteq S(A^k)$$, equality must hold. Now the result follows.