Contour integration of Resolvent Let $B$ be a Banach space and $Q\colon B\to B$ be a linear operator with eigenpairs $\{(q_{k}, v_{k})\}$ with $v_{k}$ orthonormal. In this document, it is shown that if $C$ is a contour containing only the eigenvalue $q_{j}$, then
\begin{align}
P_{j} = -\frac{1}{2\pi i} \oint_{C} \frac{1}{Q-z} \, \mathrm{d}z,
\end{align}
where $P_{j}$ is the orthogonal projection operator onto $\mathrm{span}\{v_{j}\}$.
This identity is pretty obvious, but it seems like evaluation of the integral requires knowledge of both the eigenvectors and eigenvalues of $Q$ (eigenvectors requires to know what space the integral is a projection onto, the eigenvalues to understand where the poles are). 
However, if I know the eigenpairs of a linear operator, I know pretty much everything I need to know about it. So my questions are 1) is this identity valuable even if I know the eigenpairs of $Q$?, and 2) is there a way to evaluate the integral given a linear operator and a contour without knowing the eigenpairs?
 A: You have stated that you are working in a Banach space, but your terminology is not valid except for a Hilbert space. There is no notion of orthogonality of a projection or orthogonality of individual eigenvectors in a general Banach space. Furthermore, even in a Hilbert space, $P_{j}$ is not generally orthogonal, unless the operator is normal or selfadjoint. In infinite-dimensional spaces, you can have an isolated singularity of the resolvent at $q_{j}$ with $\mathcal{N}(Q-q_{j}I)=\{0\}$; that is, an isolated point of the spectrum does not imply that there are eigenvectors associated with eigenvalue $q_{j}$. There are quasinilpotent operators $Q$ with $\sigma(Q)=\{0\}$ and yet $\mathcal{N}(Q)=\{0\}$; in this case the resolvent $(Q-\lambda I)^{-1}$ has an essential singularity at $\lambda=0$. So there are no guarantees that an isolated singularity of the resolvent (i.e., an isolated point of the spectrum) is an eigenvalue. If you know that $x$ is an eigenvector with eigenvalue $q_{j}$, then $P_{j}x=x$ will be true; however $P_{j}x=x$ does not mean that $x$ is an eigenvector of $Q$ with eigenvalue $q_{j}$ because $\mathcal{N}(Q-q_{j}I)=\{0\}$ can occur for general operators, even though $P_{j}\ne 0$. This is something peculiar to infinite-dimensional spaces.
For a Banach space with an isolated point of the spectrum at $q_{j}$, the operator formed as the resolvent integral over a contour enclosing $q_{j}$ but no other point of the spectrum is a projection $P_{j}$. That is $P_{j}^{2}=P_{j} \ne 0$. And, $P_{j}$ commutes with $Q$ for obvious reasons. But you can't say much more than that. The space onto which $P_{j}$ projects may be a cyclic type of space that is found in Jordan canonical form, or something more general in infinite dimensional space, including $P_{j}=I$ even though $Q\ne 0$. Because of the functional calculus,
$$
         (Q-q_{j}I)^{n}P_{j} = -\frac{1}{2\pi i}\oint_{C}(z-q_{j})^{n}(Q-z I)^{-1}\,dz
$$
When you assume that $q_{j}$ is an isolated singularity of the resolvent (as is being discussed here,) it could be a pole or an essential singularity. It is a pole of order $n$ iff
$$
                    (Q-q_{j}I)^{n}P_{j}=0,\;\;\;(Q-q_{j}I)^{n-1}P_{j} \ne 0.
$$
The above condition is what you see for a Jordan block: $QP_{j}=q_{j}P_{j}+NP_{j}$ where $N$ is nilpotent of order $n$ in the case of a pole for the resolvent of order $n$. In infinite dimensions, the singularity of the resolvent may be an essential singularity, which doesn't make it easy to say much about $QP_{j}$; you may not have eigenvectors at all for the case of an essential singularity at $q_{j}$. In the case of a pole of the resolvent, you will have some eigenvectors, just as for the Jordan canonical form, because $(Q-q_{j}I)=0$ on $\mathcal{R}(Q-q_{j}I)^{n-1} \ne \{0\}$. You have a pure eigenspace at an isolated point of the spectrum, $q_{j}$, iff the resolvent has a pole of order $1$ at $q_{j}$.
In a Hilbert space, normal operators and selfadjoint operators cannot have higher order poles. This is because $Q^{2}x=0$ for a normal operator iff $Qx=0$. That's why any spectral decomposition in that case doesn't require the cyclic subspaces you see associated with general Jordan forms (or worse in infinite dimensions.) Furthermore, any isolated singularity does give rise to an orthogonal projection $P_{j}$. However, for a general $Q$ on a Hilbert space, the operator $P_{j}$ may not be an orthogonal projection, and won't be unless the range of $P_{j}$ and its orthogonal complement are invariant under $Q$, which is not guaranteed for a non-normal $Q$. So the expected property of $P_{j}$ is that it is not going to be an orthogonal projection for a general $Q$ on a Hilbert space; of course the whole concept cannot be discussed at all for a general Banach space where there is no notion of orthogonality or orthogonal projection.
Hope this helps.
