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

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