Wikipedia say the following about compact operators on a Hilbert space:

Corollary For every compact self-adjoint operator $T$ on a complex separable infinite-dimensional Hilbert space $H$, there exists a countably infinite orthonormal basis ${f_n}$ of $H$ consisting of eigenvectors of $T$, with corresponding eigenvalues ${\mu_n} \subseteq \mathbb{R}$, such that $\mu_n \to 0$, as $n \to \infty$.

Under the assumption that $H$ is separable, is this an alternative characterization of self-adjoint compact operators?

  • $\begingroup$ Yes, every such operator is compact and self-adjoint. $\endgroup$ – Qiaochu Yuan Sep 25 '17 at 19:01
  • $\begingroup$ If I generalize to not necessarily self adjoint is this equivalent to generalizing to the complex eigenvalue setting? $\endgroup$ – Alesandro Levi Sep 25 '17 at 20:26
  • $\begingroup$ Also, if $K(H)$ is the symbol for the compact operators, what is the symbol for the operators which are both self-adjoint and compact? $\endgroup$ – Alesandro Levi Sep 25 '17 at 20:30
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    $\begingroup$ No, generalizing to complex eigenvalues gets you the normal compact operators. Compact operators that aren't normal won't have an orthonormal basis of eigenvectors, although they will still have a singular value decomposition. $\endgroup$ – Qiaochu Yuan Sep 26 '17 at 3:49

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