This question should be standard, but I didn't find it in the books. For a compact operator $T$ on a Hilbert space $H$, we know that every spectral value $\ne 0$ is an eigenvalue, that each generalised eigenspace $$ H(\lambda)=\bigcup_{n\ge 1}\ker(T\lambda)^n, \ \ \ \lambda\ne 0 $$ is finite dimensional and that the eigenvalues can only accumulate at zero. Is it true that $\bigoplus _{\lambda\in\mathbb C}H(\lambda)$ is dense in $H$? Here $\lambda=0$ is included in which case the gerenalised eigenspace may be infinitedimensional.
1 Answer
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No.
Consider the Volterra operator $V: L^2(0,1) \to L^2(0,1)$ which is given by $(Vf)(x) = \int_0^x f(y) \, \mathrm{d}y$. This operator is compact and does not have any eigenvalues, so every generalized eigenspace is $\{0\}$.

1$\begingroup$ You probably want to mention that it is a compact operator. $\endgroup$– YCorDec 6, 2023 at 11:50