Eigenvalues of projector matrix

The spectrum of a projection is contained in $\{0, 1\}$, as $$(\lambda I-P)^{-1} = \frac{1}{\lambda} ( I - P) +\frac{1}{\lambda-1 }P$$Only $0$ and $1$ can be an eigenvalue of a projection, the corresponding eigenspaces are the range and kernel of the projection. wikipedia

What is the above equation called? How is it obtained?

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It seems more convincing to me to note that we have $P^2 = P$, so $P$ satisfies $X(X - 1)$. – Dylan Moreland Jan 4 '12 at 14:46
Yeah. But I really want to know what that equation is? – Pratik Deoghare Jan 4 '12 at 16:05
If you want a name, it is the "partial fraction decomposition of the resolvent" of $P$. – Mariano Suárez-Alvarez Jan 4 '12 at 18:36

Multiply the right side of your equation with $(\lambda I-P)$. Using $P^2=P$ you can verify that the product is $I$ for all $\lambda\ne 0,1$. It follows that $\lambda I-P$ is nonsingular for all $\lambda\ne 0,1$; therefore $0$ and $1$ are the only possible eigenvalues.
I don't know whether this equation has a name. You can "reinvent" it by looking at a two-dimensional situation where $P: (x,y)\mapsto (x,0)$ is the projection onto the $x$-axis. In this case $I-P$ is the projection onto the $y$-axis. You then have to answer the following question: How can I get back the point $z:=(x,y)$ when I'm given the point $w:=(\lambda I-P)z=(\lambda x-x,\lambda y)$?
One-line alternative solution: If $P$ is a projection operator, then $P^2 = P$ by defintion. Thus in particular, the eigenvalues of $P^2$ and $P$ coincide. But by the Spectral Mapping Theorem, the eigenvalues of $P^2$ are: $\gamma^2$ s.t. $\gamma$ is an eigenvalue of $P$. Thus it must hold that $\gamma^2 = \gamma$ for each eigenvalue of $P$, and so the eigvenvalues of $P$ are contained in {0, 1}.