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by dataflow
1040 days ago
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Thanks! But I think you might've missed a subtlety here: > This should generalize easily to the complex/Hermitian case. This doesn't seem to be true, in that it's actually impossible to have a non-Hermitian matrix C such that x†Cx > 0 over the complex numbers for all x. Whereas over the real numbers, with a matrix R, you can have x'Rx > 0 such that R is asymmetric. The subtlety here is that x itself can be complex in the complex case, which further constraints C to be Hermitian - see the Wikipedia link I posted above. In other words, "complex definiteness" is actually a stronger condition than "real definiteness", even for matrices without an imaginary part. |
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Nice catch, it's been a few years since I had to think about these details.