Normal endomorphism is unitary diagonalizable Let $V$ be some finite-dimensional unitary vector space and $f\in L(V,V)$ be some normal endomorphism. Show that $f$ is unitary diagonalizable.
Hint: Use the following proposition: If $f_1,f_2\in L(V,V)$ are commuting endomorphisms, then $f_1$ and $f_2$ are simultaneously diagonalizable.

I think, what I have to show is that
$$
UM(f;b)U^*=D
$$
where $M(f;b)$ is the transformation matrix corresponding to $f$ (with respect to some basis $b$), and $U$ is an unitary matrix, and $D$ is a diagonal matrix.
I am not sure if I am allowed to use the spectral theorem for normal endomorphisms which tells me that, since $f$ is normal, there is an orthonormal basis $e$ consisting of eigenfunctions of $f$ such that the corresponding $M(f;e)$ is diagonal.
 A: The precise approach here depends on what exactly your definition of "unitarily diagonalizable" is; it's clear what the definition should be for a complex matrix, but it's not clear what the definition should be for a transformation over an arbitrary (finite-dimensional) unitary space.
What I suspect is meant is the following: $f$ is unitarily diagonalizable iff there exists a unitary transformation $U:\Bbb C^n \to V$ (where $n = \dim(V)$) such that $U^* \circ f \circ U$ (a linear map over $\Bbb C^n$) is a diagonal matrix (or more precisely, corresponds to multiplication by a diagonal matrix). With that in mind, let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $f$, and $v_1,\dots,v_n$ a corresponding orthonormal set of eigenvectors. Show that the map
$$
U(x_1,\dots,x_n) = x_1 v_1 + \cdots+ x_n  v_n
$$
has this property. It is helpful to note/show that its adjoint has the form
$$
U^*(v) = (\langle v,v_1\rangle , \dots, \langle v,v_n\rangle).
$$

Perhaps the definition that you would prefer is that $f$ is unitarily diagonalizable iff for an orthonormal basis $\mathcal B$ (i.e. for at least one such basis), the matrix $[f]_{\mathcal B} = [f]^{\mathcal B}_{\mathcal B}$ (of $f$ relative to $\mathcal B$) is diagonalizable.
We can show that this holds with the help of the proof outlined above. Let $\mathcal A$ denote the standard basis of $\Bbb C^n$. Let $D$ denote the diagonal matrix corresponding to $U^* \circ f \circ U$ (i.e. $D = \operatorname{diag}(\lambda_1,\dots, \lambda_n)$). We can argue that
$$
[U^* \circ f \circ U]^{\mathcal A}_{\mathcal A} = D \implies\\
[U^*]_{\mathcal A}^{\mathcal B} [f]^{\mathcal B}_{\mathcal B} [U]^{\mathcal A}_{\mathcal B} = D \implies\\
[U]_{\mathcal B}^{\mathcal A *} [f]_{\mathcal B} [U]^{\mathcal A}_{\mathcal B} = D.
$$
So, $[f]_{\mathcal B}$ is indeed unitarily diagonalizable with the unitary change of basis matrix $[U]^{\mathcal A}_{\mathcal B}$.
Note: $[U]^{\mathcal A}_{\mathcal B}$ denotes the matrix of a transformation relative two separate bases, $\mathcal A$ being the basis of the domain and $\mathcal B$ the basis of the codomain.

All that said, if it really is enough to find one orthonormal basis $\mathcal B$ such that $[f]_{\mathcal B}$ is diagonalizable, we might as well choose the basis that diagonalizes $f$, i.e. it basis of eigenvectors. In particular, if $\mathcal B = \{v_1,v_2,\dots,v_n\}$, then we simply have $[f]_{\mathcal B} = D$.
