Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $S((\ker S)^{\perp})$?

share|cite|improve this question
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,

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.