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?
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?
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.
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 $(\Leftarrow)$?