# Characteristic Polynomial of a Matrix $V$.

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 $V$?

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

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 $$\prod_{i=1}^m (x - f(\lambda_i))^{k_i}$$
• 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. – Paul Feb 27 '15 at 18:39
• If we know all the eigenvalues of $f(V)$ up to multiplicity, then we can find its characteristic polynomial. – Omnomnomnom Feb 27 '15 at 20:01