Riesz projector

Remarks

The integral does not depend on the choice of $\Gamma$, by the Cauchy theorem for analytic operator-valued functions. The operator $P = P_{\sigma_1}$ is a projection, $P^2 = P$, and it commutes with $T$. So $X = PX \oplus (I - P)X$ with both summands closed and invariant under $T$, and the restrictions of $T$ have spectra $\sigma_1$ and $\sigma_2$.

Example: let $T = \operatorname{diag}(1, 2)$ on $\C^2$ and $\sigma_1 = \set{1}$, with $\Gamma$ the circle $\abs{\lambda - 1} = \tfrac12$. Then $(\lambda I - T)^{-1} = \operatorname{diag}(1/(\lambda - 1), 1/(\lambda - 2))$, and Cauchy's integral formula gives $P_{\sigma_1} = \operatorname{diag}(1, 0)$.

Non-example: for the right shift on $\ell^2$, the spectrum is the closed unit disc, which is connected. A subset such as the closed disc of radius $\tfrac12$ is not separated from the rest of the spectrum, so it has no Riesz projector. Only $\sigma_1 = \emptyset$ and $\sigma_1 = \sigma(T)$ are allowed, giving $0$ and $I$.