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.
Best Answer
$(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.