# isolated point of spectrum of compact self-adjoint linear operator on infinite-dimensional separable Hilbert space.

Let $$H$$ be an infinite-dimensional separable Hilbert space over $$\mathbb{R}$$, and let $$K : H \to H$$ be a compact self-adjoint linear operator. Prove that if $$0$$ is an isolated point of the spectrum of $$K$$, then $$0$$ is an eigenvalue of $$K$$ with infinite-dimensional eigenspace.

My attempt: since $$K$$ is compact operator on an infinite dimensional hilber space we have that $$0\in \sigma(K)$$ and $$\sigma(K)=\sigma_p(K) \cup\{0\}$$. Suppose that $$0 \notin \sigma_p(K)$$, then it must be that there exists some sequence $$(\lambda_j)_{j\ge 1}\in \sigma_p(K)$$ such that $$\lim_{j\to \infty}\lambda_j = 0$$. But since $$0$$ is an isolated point of the spectrum , such sequence cannot exists. Hence , $$0\in \sigma_p(K)$$ .

Also, Since $$H$$ is separable let $$(e_n)_{n\ge 1}$$ be the orthonormal basis then since $$\lambda =0$$ is an eigenvalue of $$K$$, we have that $$\forall e_n \implies K(e_n)=0(e_n)$$, so all basis vectors (which are countably infinite) can be the eigenvectors for the eigenvalue $$0$$, so dimension of the eigenspace is also infinite.

• I don't see anything wrong. Is there some point that you doubt? – Disintegrating By Parts Apr 8 at 4:33

Since $$0$$ is an isolated element,the orthogonal space of $$Ker T$$ cannot be infinite dimensional, since from the spectral theorem, in this situation the orthogonal space of $$Ker T$$ would have an orthogonal basis $$e_i$$ with $$T(e_i)=c_ie_i, lim_nc_n=0$$ this is in contradiction with the fact that $$0$$ is isolated, thus the spectral theorem implies that the orthogonal of $$Ker T$$ is finite dimensional and $$Ker T$$ infinite dimensional.