A+c_1B, …, A+c_{n+1}B nilpotent implies A and B nilpotent

Posted: February 14, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem. Let $A$ and $B$ be $n \times n$ matrices with entries from some field $k.$ Let $c_1, \cdots , c_{n+1}$ be $n+1$ distinct elements  in $k$ such that $A+c_1B, \cdots , A+c_{n+1}B$ are all nilpotent. Prove that $A$ and $B$ are nilpotent.

Solution. Since $A + c_1B, \cdots , A+c_{n+1}B$ are nilpotent, their eigenvalues are all zero and hence, by the Cayley-Hamilton theorem, $(A+c_iB)^n=0,$ for all $1 \leq i \leq n+1.$ Let $x$ be an indeterminate. So the equation

$(A+xB)^n = 0 \ \ \ \ \ \ \ \ \ \ \ \ (1)$

has at least $n+1$ roots $x =c_1, \cdots , c_{n+1}$ in $k.$ We will show that this is not possible unless $A^n=B^n=0,$ which will complete the solution.  To do so, let’s expand the left hand side in $(1)$ to get

$B^nx^n + D_1x^{n-1} + \cdots + D_{n-1}x + A^n = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ (2),$

where each $D_i$ is in the $k$-algebra generated by $A$ and $B.$ Let $1 \leq i,j \leq n$ and let $a_{ij}$ and $b_{ij}$ be the $(i,j)$-entries of $A^n$ and $B^n$ respectively. Then the $(i,j)$-entry of the matrix on the left hand side of $(2)$ is

$b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij}=0, \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

where $d_1, \cdots , d_{n-1}$ are the $(i,j)$-entries of $D_1, \cdots , D_{n-1}.$ So $(3)$ is telling us that the polynomial $p(x) = b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij} \in k[x],$ which has degree at most $n,$ has at least $n+1$ roots in $k.$ This is not possible unless $p(x)=0.$ Thus $a_{ij}=b_{ij}=0. \ \Box$