Dimension of operators Question: Let $T: \ell^2 \to \ell^2$ be self-adjoint and compact. For $\lambda \in \mathbb{R}$, let
$$
S_\lambda = \overline{\text{Span} \{ v \in \ell^2 \mid Tv = \gamma v \text{ for some } \gamma \le \lambda \}}
$$
Show that $T$ maps $(S_1)^\perp$ to itself. In general, for what $\lambda$ are $S_\lambda$ and $(S_\lambda)^\perp$ finite dimensional and infinite dimensional?
Comments: This certainly involves the spectral theorem, but I'm struggling to put together a solution.
Any help is appreciated.
 A: As you say, the spectral theorem can help. Since $T$ is compact and selfadjoint, there exists an orthonormal basis $\{e_n\}_{n\in\mathbb N}$ such that 
$$
T=\sum_{k=0}^\infty \gamma_n\,P_n,
$$
where $P_n=\langle\cdot,e_n\rangle e_n$ and $\gamma_n\to0$.
If follows that
$$
S_\gamma=\overline{\text{span}\,\left\{ e_n:\ \gamma_n\leq\lambda\right\}}.
$$
Then 
$$
S_\lambda^\perp=\overline{\text{span}\,\left\{e_n:\ \gamma_n>\lambda\right\}}
$$
As $Te_n=\gamma_ne_n\in\text{span}\,\{e_n\}$ for all $n$, we have that $S_\lambda$ and $S_\lambda^\perp$ are invariant for $T$ for all $\gamma\in\mathbb R$. 
As for dimensionality, if $\lambda>0$, then $\dim S_\lambda=\infty$ (because either $\ker T$ is infinite-dimensional or there exist infinitely $\gamma_n$ with $\gamma_n<\lambda$) and $S_\lambda^\perp$ is finite dimensional (because there are finite many positive eigenvalues, each finite-dimensional). The opposite happens when $\lambda<0$: in that case, $S_\lambda$ is finite-dimensional and $S_\lambda^\perp$ is infinite-dimensional. 
When $\lambda=0$, all possibilities can arise. For example,


*

*if $T=\sum_{n=1}^\infty \frac1n\,P_n$, then $S_0=\{0\}$ and $S_0^\perp=\ell^2$. 

*if $T=-\sum_{n=1}^\infty \frac1n\,P_n$, then $S_0=\ell^2$ and $S_0^\perp=\{0\}$. 

*if $T=\sum_{n=1}^\infty \frac1n\,P_{2n}$, then both $S_0$ and $S_0^\perp$ are infinite-dimensional. 

