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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](http://en.wikipedia.org/wiki/Fredholm_alternative) see also [here](http://www.math.ksu.edu/~ramm/papers/419amm.pdf).2012-01-06

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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.2012-01-06
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Since $T$ is injective, then you write $T=-\lambda(I-\dfrac{1}{\lambda}T)$ since $\dfrac{1}{\lambda}T$ is compact then $I-\dfrac{1}{\lambda}T$ is invertible and therefore surjective.


Remember that for $T\colon E\to E$ compact and E banach if $(I-T)$ is inyective then $(I-T)$ is inversible. That's all my work, do you know how to prove $(\Leftarrow)$?