Normal operator $f \in L(V,V)$ adjoint as a polynomial in $f, f^*=p(f)$.

I'm preparing for a Linear Algebra exam, grad school level.

If $V$ is a complex vector space "unitaire" (term in French, but I can't find this term anywhere except in my class notes, I think it's "hermitien", meaning a vector space with a scalar product, i.e. complex prehilbert space), and $f$ a normal endomorphism of $V$, I need to show there is a polynomial $P\in\Bbb C[X]$ such that $f^*=P(f)$.

I think it's related to $f$ being diagonalizable, because I've found such a $P$ for $f$ and $g$ diagonalizable endomorphisms that commute. The Wikipedia site for normal operators (French version) says a normal endomorphism in a complex prehilbert space is diagonalizable in an orthonormal basis, but I don't have anything in my class notes saying that, a proposition I have requires $P_f$ (characteristic polynomial) be factorizable. Is there some property of normal endomorphisms in a prehilbert space that I'm missing? Thank you in advance.

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I will assume that $V$ has finite dimension.

Since $f$ is normal, it is diagonalizable.

So there is an orthonormal basis where $f=diag(\lambda_1,\ldots,\lambda_n)$.

By Lagrange interpolation (for instance), you can find a polynomial $P$ such that $P(\lambda_j)=\bar{\lambda_j}$ for $j=1,\ldots,n$.

Then $P(f)=f^*$.

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Yes, V has finite dimension, sorry. That's exactly what I wanted to do, but I wasn't sure f was diagonalizable. Is that because it's a complex vector space, or is that always true for f normal? –  JKH Jan 13 '13 at 21:04
@JKH The orthonormal diagonalization of a normal matrix (at least the proof I know) requires the field to be algebraically closed. –  1015 Jan 13 '13 at 21:16
Ok, that's what I was thinking, C is algebraically closed, so the characteristic polynomial is factorizable, so f is diagonalizable by the proposition I have from class. Thank you for your help. –  JKH Jan 13 '13 at 21:17
@JKH For instance, on $\mathbb{R}^2$, the matrix with first row $(0 -1)$ and second row $(1,0)$ is normal but not diagonalizable. But maybe we can still express its adjoint as a polynomial... I don't know. –  1015 Jan 13 '13 at 21:25