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Suppose that we had a matrix $V$ which was a complex square matrix($n$ x $n$) with characteristic polynomial $\prod_{i=1}^{m} (x - \lambda_i)^{k_i}$ with $f(x)$ is a complex polynomial.

What is the characteristic polynomial $f(V)$?

I'm not entirely sure what this question is asking. Because the way I look at it, if we have that $f(x)$ is the characteristic polynomial for $V$. Doesn't that imply that $f(V)$ is $0$ by the Cayley-Hamilton theorem and satifisies the equation: $(V - I_n\lambda_1)^{k_1} (V - I_n \lambda_2)^{k_2} \dots(V - I_n \lambda_m)^{k_m} = 0$ so wouldn't this be the characteristic polynomial $f(V)$ for the matrix $V$?

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  • $\begingroup$ In this question, $f$ is not necessarily the characteristic polynomial. $\endgroup$ – Omnomnomnom Feb 27 '15 at 18:22
  • $\begingroup$ @Omnomnomnom I don't understand what you mean by that. Why is that the case? $\endgroup$ – Paul Feb 27 '15 at 18:24
  • $\begingroup$ Because all they said is that "$f$ is a complex polynomial". Not every complex polynomial is the characteristic polynomial of $V$. $\endgroup$ – Omnomnomnom Feb 27 '15 at 18:25
  • $\begingroup$ Oh wait, I misread the question. $\endgroup$ – Omnomnomnom Feb 27 '15 at 18:26
  • $\begingroup$ Perhaps you're meant to (partially) prove the Cayley-Hamilton theorem, then $\endgroup$ – Omnomnomnom Feb 27 '15 at 18:26
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Hint: if $\lambda$ is an eigenvalue of $V$, then $f(\lambda)$ is an eigenvalue of $f(V)$.

You should find that the characteristic polynomial of $f(V)$ is $$ \prod_{i=1}^m (x - f(\lambda_i))^{k_i} $$

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  • $\begingroup$ I see this as $f(V)$ divides $\prod_{i=1}^{m} (x - \lambda_i)^{k_i}$ but I can't figure out how to use this hint. $\endgroup$ – Paul Feb 27 '15 at 18:39
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    $\begingroup$ If we know all the eigenvalues of $f(V)$ up to multiplicity, then we can find its characteristic polynomial. $\endgroup$ – Omnomnomnom Feb 27 '15 at 20:01

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