For a normal operator $T$ on a complex ips, let $p \in \mathcal{P}(\mathbb{C})$, and prove there exists $S$ such that $p(S) = T$

Question

PROBLEM: Suppose $T$ is a normal operator on a finite-dimensional complex inner product space $V$, and let $p \in \mathcal{P}(\mathbb{C})$. Prove that there exists a normal operator $S \in \mathcal{L}(V)$ such that $p(S) = T$. (The hint I've been given is to define $S$ on a basis and use the Fundamental Theorem of Algebra).

MY APPROACH: I know by the spectral theorem that $T$ has an orthonormal basis consisting of eigenvectors. I'm not sure if the orthonormality of the basis is important, but the basis of eigenvectors implies that $T$ is diagonalizable and that $V = E(\lambda_1, T) \oplus ... \oplus E(\lambda_m, T)$. I'm wondering if I should think about the restriction of $T$ to each of those eigenspaces? But that doesn't seem to lead anywhere...

Answer 1

I'm assuming $\mathcal{P}(C)$ is the set of polynomials over the complex numbers? If so, then just define $S$ on the eigenvectors of $T$ and extend linearly.

That is, if $Tv = \lambda v$, then define $Sv = \tau v$ where $p(\tau) = \lambda$ and $\tau$ exists by the fundamental theorem of algebra.