Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hey I'm stuck on this question, I'll be glad to get some help.

$A$ is a matrix, $f(x)$ is a polynomial such that $f(A)=0$. Show that every eigenvalue of $A$ is a root of $f$.

Well, I thought of something but I got stuck: we know that if $t$ is an eigenvalue of $A$, then $f(t)$ is an eigenvalue of $f(A)$, so letting $v$ be an eigenvector for $t$: $$f(A)=0\implies f(t)v=f(A)v=0\implies (v\ne 0)\implies f(t)=0$$ although I think that the last step is not true. Any help?


share|cite|improve this question
Yes..I thought about it but didn't know how to use it here – user137191 Mar 22 '14 at 11:28
you basically did right: $f(t)v=0$ and $v\neq 0$ implies $f(t)=0$. – user126154 Mar 22 '14 at 11:29
But suppose v is a matrix B. if AB=0 it doesn't require that A=0 or B=0. Isn't it the same case ? – user137191 Mar 22 '14 at 11:31
Here $v \ne 0$ is a scalar! – user119228 Mar 22 '14 at 11:31
@Julien I think you mean $f(t)$ is a scalar... – fgp Mar 22 '14 at 11:33

Assuming $A$ is diagonalizable, you can write it as $A=PDP^{-1}$ and transcribe your equation into

$f(A)=P( \rm{diag\,}{f(\lambda)}) P^{-1}$

If $f(A)$ is to be zero, and $P$ is nonsingular, then the diagonal matrix on the right must be zero, which explicitly states that all the eigenvalues are roots of $f$.

However, this may assume too much. Depending on how rigorous a proof you need.

share|cite|improve this answer
The OP proof is rigorous and correct. – user119228 Mar 22 '14 at 11:36
I don't know that A is diagonalizable though.. But thankyou :) – user137191 Mar 22 '14 at 11:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.