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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.

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    $\begingroup$ I don't see anything wrong. Is there some point that you doubt? $\endgroup$ – COVID-20 Apr 8 at 4:33
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This just a way to write the spectral theorem:

https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Spectral_theorem

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.

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