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

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

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 $$B^n=O$$.
• 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 7 at 8:07
• Do we even need the fact that $A$ is nilpotent tho? – EuxhenH Jan 7 at 8:12
• @EuxhenH Maybe we can assume that one of $c_k$ is 0. – Song Jan 7 at 8:13
• @Song That would mean that $A$ is nilpotent as a consequence, but we still don't need it in the problem statement. – EuxhenH Jan 7 at 8:16