Normal operators are unitarily equivalent iff they have the same spectrum

Question

Here is a more general claim that implies the argument in the title:
Show that if $S$ and $T$ are two normal operators, then there is a $*$-isomorphism $\pi: C^*(S)\to C^*(T)$ s.t. $\pi(S)=T$ iff $\sigma(S)=\sigma(T)$.

$\Rightarrow$ Suppose $\pi$ is such an isomorphism. I know that if $\phi : A\to B$ injective $*$-homomorphism then $\sigma(a)=\sigma(\phi(a))$. As $\pi$ is an isomorphism $\sigma(S)=\sigma(\pi(S))=\sigma(T)$.

$\Leftarrow$ By Gelfand transform there exist $*$-isomorphisms
$G_1: C(\sigma(S)) \to C^*(1,S)$ and $G_2: C^*(1,T)\to C(\sigma(T))$. Thus $H:=G_2\circ \pi \circ G_1 : C(\sigma(S)) \to C(\sigma(T))$ is a $*$-isomorphism. So, there exists $F: \sigma(T) \to \sigma(S)$ homeomorphism s.t. $H(g)=g \circ F$ for all $g\in \sigma(T)$.
But we know that $\sigma(T)=\sigma(S)$. If I could show that $F=id$ then I can finish the proof.

Any help for completing the proof will be appreciated.
Clearly, the argument implies that unitarily equivalent normal operators have the same spectrum.
So, I wonder what we can say about the spectrum of unitarily equivalent operators that are not necessarily normal?

Answer 1

It seems as if you confused what to assume and what to prove in the second part. If you want to show the existence of $\pi$, you cannot include it in the definition of $H$.

But if you get your arguments sorted out, it's quite obvious. Let $\pi=G_2^{-1}\circ G_1^{-1}$ and $f\colon \sigma(S)=\sigma(T)\to \mathbb{C},\,z\mapsto z$. By definition, $G_1(f)=S$ and $G_2^{-1}(f)=T$. Thus, $\pi(S)=G_2^{-1}(f)=T$.

As for your second question, conjugation by a unitary is a $\ast$-automorphisms. In particular, it preserves spectra.

Answer 2

The proof of the converse is simple: if $\sigma(T)=\sigma(S)$, then $C(\sigma(T))=C(\sigma(S))$. Then $G_2\circ G_1^{-1}$ is the $*$-isomorphism you are looking for.

Now, it is not true that the statement you gave implies unitary equivalence; you cannot ignore multiplicity. Consider $$ S=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix},\ \ \ T=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}. $$ Both have spectrum $\{0,1\}$ but they are not unitarily equivalent. Actually, they are not even similar (as you can see, for instance, since they have different trace).

In fact, things can be more drastic: you can have normals with the same spectrum in highly different environments: for instance $S=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $T=1_{2\mathbb N}\in\ell^\infty(\mathbb N)$ (i.e., an alternating sequence of ones and zeroes), both with spectrum $\{0,1\}$.