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.

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 $P(f)=f^*$.
