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Is the spectral theorem for self-adjoint compact operators on a Hilbert space also true if the Hilbert space is real (instead of complex)?

Wikipedia says this is true.

However, it seems to me that in none of the references of that article, the theory is established for real Hilbert spaces.

If it is true I would be grateful for a reference.

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  • $\begingroup$ Actually, now I think of it, one probably does better to just follow the proof of the spectral theorem for the complex case, rather than appeal to it as a black box. If you look at how the theorem is usually proved : just like the finite-dimensional case one produces a non-zero eigenvalue (assuming your operator is non-zero) and uses self-adjointness to prove that the eigenvalue is real & that the perp of the eigenspace is invariant under your operator. Then you induct, using compactness to show that your sequence of eigenspaces (together with the kernel) eventually exhausts the whole space. $\endgroup$
    – Yemon Choi
    Commented Apr 25, 2020 at 20:35
  • $\begingroup$ Now in the sketch above, the only place where I relied on having complex scalars was to produce a non-zero eigenvalue; everything else works for real inner products in exactly the same way as for complex inner products. So you are in business as soon as you can prove that a self-adjoint compact operator $T$ on a real Hilbert space always has a (real) eigenvalue. But there are (standard) arguments which show that at least one of $\Vert T\Vert$ or $-\Vert T\Vert$ will be an eigenvalue, and these don't rely on complex-variable techniques $\endgroup$
    – Yemon Choi
    Commented Apr 25, 2020 at 20:38
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    $\begingroup$ Answered in greater generality here. $\endgroup$
    – Nik Weaver
    Commented Apr 25, 2020 at 20:46
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    $\begingroup$ Yes, Raleigh-Ritz (sp?) method of proving existence of an eigenvalue/eigenvector is an essentially "real" method, anyway. $\endgroup$
    – paul garrett
    Commented Apr 25, 2020 at 21:40

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