On the spectral decomposition for compact operators in a Banach space

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
$$
P_{lambda_j}=frac{1}{2pi i}int_{gamma_j} (z-T)^{-1} dz,
$$
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.
$$
lim_{ntoinfty}left|P-sum_{j=1}^nP_{lambda_j}right|=0.
$$

Then two points must be addressed:

1. Is $Q$ always a well-defined projector?
2. 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 $0insigma(TQ)$ and any other element of the spectrum is an eigenvalue.
We assume $Qneq0$, 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?