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- Thread starter LarryC
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bhobba

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An example of C is take a Hilbert Space and the identity operator. All vectors have the same eigenvalue ie 1 and any orthogonal infinite basis generates a subspace with the same eigenvalue. If this is beginning level linear algebra I think the author of the problem needs to think a bit harder about appropriate questions to ask and its wording. I know finite dimensional linear algebra well, Hilbert Spaces, and even Rigged Hilbert Spaces and would never ask a question like that.

My suggestion is take it along to a tutorial or see your lecturer about it.

Thanks

Bill

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vela

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Show us your proof for part (b). There's probably a step in there that doesn't necessarily hold for infinite dimension subspaces.(a) and (b) are fairly traditional, but I have trouble understanding the phrasing of (c). What makes the infinite dimensionality in (c) different from (a) and (b)?

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bhobba

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Show us your proof for part (b). There's probably a step in there that doesn't necessarily hold for infinite dimension subspaces.

Yes B is not that hard for finite dimensional spaces - as a hint just look into the spectral theorem. I can write the proof in a few lines using bra-ket notation. Extending that to infinite dimensional spaces is much more difficult (it requires the so called Nuclear Spectral theorem - also called the Gelfand-Maurin theorem) which may be the purpose of the question - infinite dimensional spaces are problematical.

Added later - if you are still stuck then under OCW MIT has a full analysis of this problem. It must be from a standard textbook - maybe Griffiths or something like that otherwise they probably would not have posted a write up on it. So Google is your friend here - and the answer is not 100^% what the question would imply - but its in your court to now. If gets around the infinite dimensional case by using infinite matrices - sneaky - but maybe not quite rigorous.

Thanks

Bill

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bhobba

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Thanks

Bill

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