5
$\begingroup$

Let $A$ be a square matrix with real or complex coefficients of size $n$. Define its characteristic polynomial by $\chi_A(X) = \det(A-XI_n)$ (or $\det(XI_n-A)$ if you prefer). The question is : Prove that there exists a positive integer $q$ such that $\chi_A^q(A)=0$ where $\chi_A^q(X)= \chi_A(X)\times\cdots\times \chi_A(X)$ without using the Cayley-Hamilton theorem or the definition of an ideal of a ring.

My problem with this problem is to find a "good" proof that doesn't mimic a proof of Cayley-Hamilton theorem. I know that it is easy to see (with a dimensional argument) that there exists a annihilating polynomial. But here I can't think of anything.

An attempt of solution in the real or complex case :

We know that there exists an annihilating polynomial (since $(I_n,A,A^2,\cdots,A^{n^2})$ can't be linearly independent, where $I_n$ is the identity matrix), we denote by $P$ such a polynomial. Then $P=\prod_{i=1}^p \left( X-\lambda_i\right)^{\alpha_i}$ ($\lambda_i\in \mathbb{C}$). We can assume that the $\lambda_i$ are roots of the characteristic polynomial. If not, then by definition of the characteristic polynomial $A-\lambda_iI_n$ is invertible. We can multiply $P$ by $(A-\lambda_iI_n)^{-\alpha_i}$, we still get an annihilating polynomial, and since we can permute all the terms, the term $\left( X-\lambda_i\right)^{\alpha_i}$ disappears. So if we take a sufficiently large power of the characteristic polynomial of $A$, we recover all the factors of $P$ times an another polynomial. This gives the desired result.

  1. What do you think of my attempt?
  2. Any ideas for the case where the coefficients of $A$ belong to an arbitrary abelian ring?
$\endgroup$
  • $\begingroup$ "real or complex" is a dead giveaway -- you should use eigenvectors. $\endgroup$ – darij grinberg May 14 '14 at 7:36
  • $\begingroup$ @darijgrinberg Can you please develop a little more? $\endgroup$ – user37238 May 14 '14 at 7:46
  • $\begingroup$ Do you know what a generalized eigenspace is? Try to show that every generalized eigenspace of $A$ is annihilated by some power of $\chi_A\left(A\right)$. $\endgroup$ – darij grinberg May 14 '14 at 23:10
  • $\begingroup$ @darijgrinberg Thank you for the hint, I'll try that. $\endgroup$ – user37238 May 15 '14 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.