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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?


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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.

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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

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