Show that if $Sp(f)=\{\lambda_1,...,\lambda_n\}$ then $Sp(Q(f))=\{Q(\lambda_1), ..., Q(\lambda_n)\}$

Question

Preparation for an exam, this question was in a test from previous year.

Given $f: E\rightarrow E$ an endomorphism, with $E$ of dimension $n$, $Q$ any polynomial. Show that if $Sp(f)=\{\lambda_1,...,\lambda_n\}$ is the spectrum of $f$, then $Sp(Q(f))=\{Q(\lambda_1), ..., Q(\lambda_n)\}$.

I know that that if $f$ is diagonalizable then $Q(f)$ is also diagonalizable but I don't see how to use it. I feel like the proof should use similar reasoning. Any ideas?

Answer 1

Let $\lambda _i$ an eigenvalue and $v_i$ an eigenvector associated to $\lambda _i$, i.e. $$f(v_i)=\lambda_i v_i.$$ In particular $$f^p(v_i)=\lambda _i^pv_i,$$ for all $p\in \mathbb N$.

Let $Q(X)=a_mX^m+...+a_1X+a_0$ a Polynomial. Then $$Q(f)(v_i)=a_n \lambda _i^mv_i+...+a_1\lambda _iv_i+a_0v_i=Q(\lambda _i)v_i,$$

and thus, $Q(\lambda _i)\in Sp \big(Q(f)\big).$