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Let $S = I - T$ where $T$ is a compact linear operator on a Hilbert space $H$. Why is it that the range of $S$ is equal to $S((\ker S)^{\perp})$?

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up vote 3 down vote accepted

let $x \in H$. As $\ker S$ is closed, we have $H = \ker S \oplus (\ker S)^\bot$, so write $x = x_1 + x_2$ with $x_1 \in \ker S$, $x_2 \in (\ker S)^\bot$. Then \[ Sx = Sx_1 + Sx_2 = 0 + Sx_2 = Sx_2 \] So we have $Sx \in S\bigl((\ker S)^\bot\bigr)$ for every $x \in H$.

As you can see, this relation doesn't depend on the fact, that your $S$ is a compact pertubation of the identity, it holds true for every $S \in L(H)$.

Hope this helps,

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