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

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...

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.
• What does $\tau$ represent here? Aug 15, 2019 at 19:25
• It's any complex number that satisfies $p(\tau) = \lambda.$ You should do this for each $\lambda = \lambda_i$ to define $S$ on the eigenbasis corresponding to $T.$ Aug 15, 2019 at 19:49