Let $T$ be a compact operator in a Banach space, then the spectrum of $T$ contains $0$ and a sequence (either finite or infinite) of eigenvalues $\lambda_1>\lambda_2>\ldots$, and $0$ is the only possible point of accumulation. For every $\lambda_j$ we can construct the spectral projector $P_{\lambda_j}$ by $$ P_{\lambda_j}=\frac{1}{2\pi i}\int_{\gamma_j} (z-T)^{-1} dz, $$ with $\lambda_j$ the only point of the spectrum enclosed by $\gamma_j$. Since $(T-\lambda_j)P_{\lambda_j}$ is nilpotent, if the number of elements in the spectrum is finite and $\lambda=0$ is also an isolated eigenvalue, then $TP_0$ is a quasinilpotent operator (perhaps nilpotent) and a Jordan-like canonical form follows. My question is whether this is true in any case. Specifically, define the operator $$ Q:=\mathbb{I}-P, $$ where, if $T$ has a finite number of eigenvalues, the projector $P$ is given by $$ P=\sum_{j}P_{\lambda_j}\ $$ and, if the number of eigenvalues of $T$ is infinite, $P$ is defined by some norm or strong limit, e.g. $$ \lim_{n\to\infty}\left\|P-\sum_{j=1}^nP_{\lambda_j}\right\|=0. $$
Then two points must be addressed:
- Is $Q$ always a well-defined projector?
- In the affirmative case, is $TQ$ a quasinilpotent operator?
Attempt
Consider the case of a finite number of eigenvalues. Then $Q$ exists and it is bounded, so that $TQ$ is compact, which means that $0\in\sigma(TQ)$ and any other element of the spectrum is an eigenvalue. We assume $Q\neq0$, otherwise the statement is trivially true. If $x$ is an eigenvector, $TQx=\lambda x$, applying $Q$ on the left, $TQx=\lambda Qx$ as $T$ and $Q$ commute each other. Therefore, $Qx$ is an eigenvector of $T$. However, this is not possible because $PQ=0$. Hence, $\sigma(TQ)=\{0\}$.
I'm not sure about the case with an infinite number of eigenvalues. Any idea?