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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.

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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.

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