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On an infinite dimensional space, the spectral theorem for compact normal operators says that the eigenvectors form an orthonormal basis which, from wikipedia, is equivalent to the space being seperable. So is the decomposition only valid for seperable spaces?

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    $\begingroup$ It is also valid in the non-separable case but in the rather trivial sense that the space splits into a direct sum of two spaces, one on which the operator vanishes and a separable one which is spanned by the eigenvectors. $\endgroup$
    – alpha
    Commented Feb 7, 2014 at 10:07

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If $T:H\to H$ is a normal operator, $\operatorname{ker}T=\operatorname{ker}T^*$ (recall that $\|Tx\|^2= (T^*Tx,x)=(T T^*x,x)=\|T^*x\|^2$). So there is a $T$-invariant orthogonal decomposition $H=\operatorname{ker}T\oplus\overline{T(H)}$. If $T$ is compact, the latter factor is separable.

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  • $\begingroup$ (this was meant to be a comment, with too much TeX) $\endgroup$
    – Pietro Majer
    Commented Feb 7, 2014 at 10:13

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