Of course, if you want to raise the eigenvalues to the power of their algebraic multiplicities, then you do get the determinant. That's another nice way of thinking about it. But it wasn't what my tweet was referring too (geometric multiplicities).
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I agree. But I do have 1 question -- what does 'degenerate' mean to you? I'm used to talking about degenerate eigenvalues, when 2 or more are identical -- is that what you mean?? Or do you mean all eigenvalues identical?? Or the same as defective, missing an eigenvector? Or??
Yes, I'd usually use it to talk about the case of 2 or more identical eigenvalues. I've also heard it used - though much less frequently - for the case where the algebraic multiplicity is 2 or more. I guess in most instances, for physicists, geom mult = alg mult.
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OK. Then just to be clear, there is still a substantive disagreement. Namely, a "degenerate non-normal" matrix -- one with 2 or more identical eigenvalues -- is not necessarily missing an eigenvector. E.g. {{2, 0, 0, 0}, {0, 2, 0, 0}, {0, 0, 3, 1}, {0, 0, 0, 1}}
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That is, if the degenerate eigenvalues correspond to a "normal subspace" -- a subspace that can be spanned by orthogonal eigenvecs -- of a non-normal matrix, then matrix need not be missing an eigenvector.
End of conversation
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