# If $A \in \mathcal{L}_c(X)$ and $X$ is Banach, then $\dim \ker (\text{id}-A) < + \infty$.

Exercise :

Let $$X$$ be a Banach space and $$A \in \mathcal{L}_c(X)$$. Show that $$\dim \ker ( \text{id} - A) < + \infty$$.

Attempt/Thoughts :

The kernel of the operator $$(\text{id}-A) : X \to X$$ is : \begin{align*}\ker(\text{id}-A) &=\{x \in X : (\text{id}-A)(x) = 0 \} \\ &= \{x \in X :x-A(x) = 0 \} \\ &= \{x \in X : A(x) = x \}\end{align*} So essentialy, we are restricting $$A$$ to the kernel of $$\text{id}-A$$, as $$A\bigg|_{\ker(\text{id}-A)} \equiv x \in \ker(\text{id}-A)$$. But $$A$$ is compact, which means that if $$\ker(\text{id}-A)$$ is Banach, then it cannot be infinite dimensional, otherwise we would not be able to achieve compactness.

Am I missing something in my elaboration here or is it as straightforward ? Maybe, how would one imply that $$\ker(\text{id}-A)$$ is indeed Banach ?

Any tips, elaborations or corrections will be appreciated.

• I am not very sure, but can you try to show that the unit ball of the kernel of $I-A$ is compact? This will show that the kernel is finite dimensional. For example, if $x_n$ are in the unit ball, then they are such that $Ax_n = x_n$ and $||x_n|| \leq 1$. Now , can you show that $x_n$ must have a convergent subsequence? Maybe using the compactness of $A$, noting it takes bounded sets to relatively compact sets? – астон вілла олоф мэллбэрг Mar 13 at 16:27
• @астонвіллаолофмэллбэрг Sure, that could be a more rigorous way. But, in my approach, is something wrong ? I mean, it seems straightforward otherwise it wouldn't be compact. – Rebellos Mar 13 at 16:28
• Yes , your idea is right. More precisely, on the kernel $A$ behaves the same as the identity, and the identity map is not compact unless the space is finite dimensional, so you are ok. The kernel of a bounded operator is closed, so automatically the space $\ker(I- A)$ is definitely Banach – астон вілла олоф мэллбэрг Mar 13 at 16:31
• Thanks for the check ! I assume I will add the sequential explanation proposed to be more rigorous. – Rebellos Mar 13 at 16:32
• Right. I don't know if you've seen it yet, but read up the spectral theory of compact operators, including the big theorem, because the finite dimensionality of this kernel is one of a few amazing things you can say about compact operators. – астон вілла олоф мэллбэрг Mar 13 at 16:41