Let $A\colon \text{dom}(A)\subset H \to H$ ($H$ is a Hilbert space) be a skew-adjoint (i.e. $A^{*}=-A$), closed and densely defined operator. Then the essential spectrum is the set of spectral values $\lambda$ for which $\lambda-A$ is not Fredholm of index zero.
For skew-adjoint operators the spectrum is a subset of the imaginary axis and hence the essential spectrum is a subset of the imaginary axis.
$\textbf{Question}$:
If we consider $\tilde{A}=A+C$ where $C$ is a compact operator,
why is the remainder of the spectrum of $\tilde{A}$ discrete and only consisting of countably many eigenvalues of finite algebraic multiplicity?
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$\begingroup$ I'm having difficulties to understand the context of the question: how do you know that the statement is true if you don't know a proof or a reference? $\endgroup$– Jochen GlueckCommented Jun 5, 2022 at 17:35
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1$\begingroup$ @JochenGlueck I came across this when I was reading arxiv.org/pdf/2112.03116.pdf ( on page 9 in the middle part) $\endgroup$– user99432Commented Jun 5, 2022 at 17:55
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1$\begingroup$ You are perhaps referring to the fact that a compact perturbation doesn't change the essential spectrum. This is known as Weyl's theorem (search for it perhaps). $\endgroup$– Christian RemlingCommented Jun 5, 2022 at 18:10
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$\begingroup$ @ChristianRemling But also the fact that the remainder of the spectrum is countable with finite algebraic multiplicity. Is there any kind of spectral theorem for Fredholm operators? A itself is not compact which is why I don't think this follows exactly from the spectral theorem for compact operators. $\endgroup$– user99432Commented Jun 5, 2022 at 18:24
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