$A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.

Question

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

The question has been asked before. – Andreas Caranti Jan 07 '19 at 09:39 — Andreas Caranti, Jan 07 '19 at 09:39

Myunghyun Song · Accepted Answer · 2019-01-07T09:00:41.280

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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 $B^n=O$.

edited Jan 07 '19 at 09:00

answered Jan 07 '19 at 08:00

Myunghyun Song

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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. – Kavi Rama Murthy Jan 07 '19 at 08:07
Do we even need the fact that $A$ is nilpotent tho? – EuxhenH Jan 07 '19 at 08:12
@EuxhenH Maybe we can assume that one of $c_k$ is 0. – Myunghyun Song Jan 07 '19 at 08:13
@Song That would mean that $A$ is nilpotent as a consequence, but we still don't need it in the problem statement. – EuxhenH Jan 07 '19 at 08:16
@KaviRamaMurthy As is always, any productive comments are welcome! and I'm totally okay with the downvote. – Myunghyun Song Jan 07 '19 at 08:16
@Song Thanks for the clarification. – Kavi Rama Murthy Jan 07 '19 at 08:18
@EuxhenH I agree wth you. The assumption that $A$ is nilpotent is not really necessary. – Myunghyun Song Jan 07 '19 at 08:20
@Song Here's a slightly different problem where $A$ being invertible might work. Given $A$ nilpotent and $\textbf{n}$ distinct $\textbf{non-zero}$ scalars $c_i$ such that $A+c_iB$ is nilpotent. Show that $B$ is nilpotent. Here one could say that the $(n+1)st$ scalar is $0$. – EuxhenH Jan 07 '19 at 08:23
@EuxhenH I guess $A$ being invertible is a typo, isn't it? Except that, I agree with you. – Myunghyun Song Jan 07 '19 at 08:53
@Song Yes, it's a typo. Thanks. – EuxhenH Jan 07 '19 at 08:57
@KaviRamaMurthy I also appreciate your comment. I should develop clearer proof writing skill. – Myunghyun Song Jan 07 '19 at 08:57

$A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.

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