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I have a Banach Space $X$ and an linear continuous operator $T\colon X\to X$ that has finite rank (i.e. $\dim {T(X)}<\infty$). Then,

$I-T$ is injective if and only if $I-T$ is surjective?

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Look up the Fredholm alternative see also here. –  t.b. Jan 6 '12 at 13:35
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1 Answer

If $T$ has finite rank, then $T$ is a compact operator. If $X$ is infinite dimensional, then the spectrum of $T$ is formed by a sequence of eigenvalues converging to zero.

One implication goes like this:

If $I-T$ is injective, then $1$ is not an eigenvalue of $T$. But the point spectrum of $T$ equals the spectrum of $T$ (perhaps without zero). Therefore $I-T$ is invertible and as a consequence, it is surjective.

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Actually, I need this to prove that the spectrum set is the same that the eigenvalues'set if T has finite rank. –  arawarep Jan 6 '12 at 14:32
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