Toeplitz operator commuting with a compact operator Let $T_\varphi$ be an analytic Toeplitz operator (meaning $\varphi \in H^\infty$). Further let $K$ be a compact operator that commutes with $T_\varphi$. Now I want to show that the spectrum of $K$ only consists of $0$.
I'm trying the following things: $T_\varphi K$ is compact so we can apply the spectral theorem for compact operators. So we know that at least $0$ is in the spectrum of this operator (not of $K$) and that the spectrum is countable with the only limit point $0$. Can I somehow use this? I also know that if $C$ is a compact operator $\|T_\varphi - C\| \geq \|T_\varphi \|$ so $\|T_\varphi(1 - K)\| = \|(1 - K)T_\varphi\| \geq \|T_\varphi \|$.
Any suggestions how I can continue?
 A: I think I got the idea:
Lemma: If $T_\varphi$ is an nonconstant analytic Toeplitz operator then the only invariant finite-dimensional subspace for $T_\varphi$ is $\{0\}$.
Proof: If $M$ is a finite dimensional subspace them $T_\varphi$ restricted to $M$ has an non-zero eigenvector (because it is compact). But this is not possible for an the Toeplitz operator on the whole space.
So now let $K$ be the compact operator that commutes with $T_\varphi$ and let $\lambda \neq 0$. Define $K_\lambda := \textrm{ker}(K - \lambda)$. Now we see that $K_\lambda$ is an invariant subspace for $T_\varphi$ as follows: $f \in T_\varphi K_\lambda$ then $f = T_\varphi  g$ for some $g \in K_\lambda$, that is $(K - \lambda)g = 0$. Because they commute we have $(K - \lambda) T_\varphi g = (K - \lambda)f = 0$, so $f \in K_\lambda$. Also note that $K_\lambda$ is finite dimensional by the spectral theorem.
Now, since $\varphi$ is nonconstant $K_\lambda = 0$. So $\lambda$ is not in the spectrum of $K$. $\{0\}$ is in the spectrum of $K$ by the spectral theorem for compact operators.
