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Let $A$ and $B$ be $n \times n$ matrices over some field with $A$ nilpotent. Now let $c_1,\ldots,c_{n+1}$ be $n+1$ distinct scalars such that $A+c_i B$ is nilpotent for all $i=1, \ldots,n+1$. Then how can I show that $B$ is also nilpotent?

Thanks

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Let us define the matrix valued polynomial $$ p(t) = (A+tB)^n = \sum_{k=0}^n C_k t^k. $$ We see that each $(i,j)$-th component $p(t)_{ij}$ of $p(t)$ is a polynomial in $t$ of degree at most $n$. By the assumption, $p(t)_{ij}=0$ has $n+1$ distinct roots $c_1,c_2,\ldots, c_{n+1}$ on the underlying field $\mathbb{F}$. (This is why: if $n\times n$ matrix $T$ is nilpotent, the it holds $T^k=O$ for some $k\le n$ and hence $T^n=O$. Thus, by the assumption, $p(c_k) = O$ for all $k=1,2,\ldots, n+1$.) This means $p(t)_{ij} \equiv 0$ and it follows that $p(t) \equiv O$. Since the coefficient $C_n$ of $t^n$ is $B^n$, it follows that $B^n=O$.

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  • $\begingroup$ Can you be more explicit. I don't see why $p(t)_{ij}=0$ has $n+1$ distinct roots. PS I saw that your answer was upvoted and then downvoted. I did not downvote. I am just seeking a clarification. $\endgroup$ – Kavi Rama Murthy Jan 7 at 8:07
  • $\begingroup$ Do we even need the fact that $A$ is nilpotent tho? $\endgroup$ – EuxhenH Jan 7 at 8:12
  • $\begingroup$ @EuxhenH Maybe we can assume that one of $c_k$ is 0. $\endgroup$ – Song Jan 7 at 8:13
  • $\begingroup$ @Song That would mean that $A$ is nilpotent as a consequence, but we still don't need it in the problem statement. $\endgroup$ – EuxhenH Jan 7 at 8:16
  • $\begingroup$ @KaviRamaMurthy As is always, any productive comments are welcome! and I'm totally okay with the downvote. $\endgroup$ – Song Jan 7 at 8:16

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