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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...

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    (This has nothing to do with k-theory, by the way.)2010-12-23
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    (You're right, but I figured that if anyone else read Atiyah's proof in his book, then they may have come across this issue as well. I'll remove the tag.) (Parenthetical conversations feel like whispering.)2010-12-23
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    I would definitively use the tag (functional-analysis)2010-12-23

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