Giving an infinite dimensional Hilbert space $H$ and a compact operator $K$ on $H$ with spectrum $\sigma \left( K\right) =\left\{ 0,\lambda _{0},\lambda _{1},\ldots \right\} $, I'm trying to prove that there is a decomposition $% M\oplus N=H$ such that $M$ and $N$ are $K$-invariant and the restrictions $% K_{M}$ and $K_{N}$ are respectively injective and quasinilpotent.
Clearly, If $N_{n}=\ker \left( E_{\lambda _{0}}\right) \cap \cdots \cap \ker \left( E_{\lambda _{n}}\right) $ for every $n\geq 0,$ where $E_{\lambda _{n}} $ is the spectral projection associated to $\lambda _{n},$ then $N_{n}$ is a closed $K$-invariant subspace and if $N=\underset{n\geq 0}{\bigcap }N_{n}$ then $N$ is a closed $K$-invariant subspace such that $K_{N}$ is quasinilpotent. On the other hand, if $M_{n}=Im\left( E_{\lambda _{0}}\right) \oplus \cdots \oplus Im\left( E_{\lambda _{n}}\right) $ for every $n\geq 0,$ then $M_{n}$ is $K$-invariant, $K_{M_{n}}$ is injective, $M_{n}\oplus N_{n}=H$, $M_{n}\cap N=\left\{ 0\right\} $ and $% M_{n}+N$ is closed since $M_{n}$ is finite dimensional. In addition; if $M=% \overline{\underset{n\geq 0}{\bigcup }M_{n}}$, then clearly $M$ is a closed $K$-invariant subspace.
I successfully proved that $M\cap N=\{0\}$ and hence $K_{M}$ is injective, but I still can not prove that $M+N=H$.
Thank you !