# Kernel of Bounded Operator on Hilbert Space

I encountered this question on an exam recently and was not able to solve it.

Suppose you have a bounded operator $$T$$ on a Hilbert space $$\mathcal{H}$$ such that $$I-T$$ is compact, where $$I$$ is the identity operator. Then $$ker(T)$$ is finite-dimensional.

I'm not really sure how to even start with this problem. We know that since $$I-T$$ is compact, the image of the unit ball is compact.

I thought it was true that $$(I-T)\mathcal{H} = I\mathcal{H} - T\mathcal{H} = \mathcal{H} - T\mathcal{H} = (T\mathcal{H})^\perp$$. However, this is not true, although I do not quite understand why.

I would be grateful for some tips on how to solve this, and maybe someone can tell me why the equation above does not hold.

$$(I-T)\mathcal{H} = I\mathcal{H} - T\mathcal{H}$$ is not correct. For example, if $$T=I$$ then $$(I-T)\mathcal{H}=\{0\}$$ but $$I\mathcal{H} - T\mathcal{H} =\mathcal{H}$$ since $$\mathcal{H}-\mathcal{H}=\mathcal{H}$$.
Let $$\{x_n\}$$ be any sequence in the unit ball of $$Ker (T)$$. Then $$(I-T)x_n=x_n$$. Since $$I-T$$ is compact there is a subsequence of $$(I-T)x_n$$ which converges. It follows that any sequence in the unit ball of $$Ker (T)$$ has a convergent subsequence. This implies that $$Ker (T)$$ is finite dimensional.