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The Hilbert–Schmidt theorem says a self-adjoint compact operator on a Hilbert space have a complete orthonormal set consisting of eigenvectors. Does that imply the space is separable?

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  • $\begingroup$ $0$ is a compact operator on any Hilbert space. A Hilbert space $H$ is separable iff there exists exists an injective compact selfadjoint operator $A$ on $H$. $\endgroup$ – DisintegratingByParts Nov 7 '14 at 18:40
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No. The eigenvectors are a complete orthonormal set for the image of the operator. The space would be separable if the operator were surjective.

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  • $\begingroup$ However, compact operators are never surjective. $\endgroup$ – Tomek Kania Nov 7 '14 at 9:28
  • $\begingroup$ They can be surjective iff the space is finite dimensional. $\endgroup$ – DisintegratingByParts Nov 7 '14 at 18:32
  • $\begingroup$ Thanks! So we can conclude that the orthogonal complement of kernel A is separable? $\endgroup$ – tomography Nov 8 '14 at 19:31

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