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I came across a line (http://www.springerlink.com/content/t523l30514754578/) about how the trace of a linear operator is not, in general, independent of the choice of orthonormal basis. The link states that such an operator may have a trace that converges for one basis but not another.

This sounds really neat but I haven't been able to find much about it, other than to learn that if you want your trace to exist and you want it independent of an orthonormal basis then you should restrict to nuclear operators.

Ultimately I'm looking for a counterexample, where the trace converges for one basis and not the other. But I'll take anything at this point.

Cheers,

Kevin

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# Nuclear operators and trace

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