# Can an accumulation point be an eigenvalue?

For an discrete (separable) infinite-dimensional Hilbert Space with a compact operator, 0 is always an accumulation point (https://www.math.ucdavis.edu/~hunter/book/ch9.pdf).

Does this mean its part of a spectrum? Also can it be an eigenvalue? Otherwise, how should I prove it is/isnt?

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This is true in every infinite dimensional Hilbert space. The spectrum is always closed, thus every accumulation point of the spectrum is an element of the spectrum. – The User May 14 '13 at 13:11
$0$ is always in the spectrum (for a compact operator on an infinite-dimensional Hilbert space). For some such operators $0$ is an eigenvalue, but for others it is not. – Gerald Edgar May 14 '13 at 14:18
Just think of diagonal operators on $l_2$, to get any sequence accumulating to 0 as spectrum, 0 being an eigenvalue or not. – Pietro Majer May 14 '13 at 18:57

The spectrum consists of all values $\lambda$ for which $T-\lambda I$ does not have a continuous inverse. Eigenvalues always lie in the spectrum, but some of the values in the spectrum may not be eigenvalues. As mentioned before, the spectrum is always closed. Of course, 0 CAN be an eigenvalue, because it is an eigenvalue for $T=0$. But it does not have to be an eigenvalue. Let $\ell^2$ be the space of all infinite sequences $(z_1,z_2,\dots)$ with $\sum_i |z_i|^2<\infty$, and define $T:\ell^2\to \ell^2$ by $T(z_1,z_2,\dots)=(z_1,z_2/2,z_3/3,\dots)$. Then $T$ is compact. The map $T$ is injective, so $0$ is not an eigenvalue. $1/n$ is an eigenvalue for all $n$, and these eigenvalues converge to 0. The element $(1,1/2,1/3,\dots)\in \ell^2$ does not lie in the image of $T$ ($(1,1,\dots)$ does not lie in $\ell^2$) so $T$ is not surjective. This means that $0$ lies in the spectrum. But all nonzero values in the spectrum of a compact operator must be eigenvalues.