why is the following thing a projection operator? Let $T: E \rightarrow E$ be an endomorphism of a finite-dimensional vector space, and let $S$ be a circle in the complex plane that does not intersect any eigenvalues of $T$. Now let $Q = \frac{1}{2\pi i} \int_S (z-T)^{-1} \, dz$. 
Why is $Q$ a projection operator?
The motivation behind this question is that the above situation occurs in a proof of Bott's periodicity theorem, but it's not clear to me that $Q$ is a projection...
 A: If $A$ is any Banach algebra (such as the algebra of endomorphisms of a finite dimensional  complex vector space), then for each subset $\Omega$ of the complex plane and each element $T$ of $A$ whose spectrum is contained in $\Omega$, holomorphic functional calculus yields a homomorphism $f\mapsto f(T)$ from the algebra of functions holomorphic in an open set containing $\Omega$ (identified if they agree on some neighborhood of $\Omega$) into $A$.  Since the function $f:(\mathbb{C}\setminus S)\to\mathbb{C}$ defined by $f(w)=\frac{1}{2\pi i}\int_S(z-w)^{-1}dz$ takes on only the values $0$ and $1$ (it gives the winding number of $S$ about $w$), $f$ is idempotent (i.e., $f(w)^2=f(w)$ for all $w\in\mathbb{C}\setminus S$), and thus $f(T)$ is an idempotent element of $A$ for each $T$ whose spectrum is disjoint from $S$.
A: Show that the integral depends continuously on $T$, and show that $Q^2=Q$ when $T$ is diagonalizable, by finding how $Q$ changes change you change $T$ by a similar matrix, and then reducing to the one dimensional case. Then use the fact that diagonalizable matrices are dense in the space of all matrices, and that $Q^2$ and $Q$ are continuous functions of $Q$.
