[dataflow]

Did positive definite not imply Hermitian? I feel like I'm forgetting my linear algebra.

Edit: Yup, Wikipedia agrees "this condition implies that M is Hermitian"; see their counterexample with a complex vector: https://en.wikipedia.org/wiki/Definite_matrix#Consistency_be...

Note: Crucially, this is specific to the field of complex numbers (hence the discussion of Hermitian vs. just symmetry). For the field of real numbers, PSD does not imply symmetry, though that's commonly assumed for convenience.

[necroforest]

kind of. you can decompose an arbitrary matrix into symmetric and antisymmetric components: R = S + A. Since A = -A^H (anti-symmetric), for any vector x, <x, Ax> = -<x, Ax> => <x, Ax> = 0. So for any matrix where <x, Rx> > 0, you can add an arbitrary anti-symmetric matrix and keep the same induced quadratic form. So people typically enforce symmetry in their definitions because it is the only part that contributes to the quadratic form and is "nicer" to work with (always diagonalizable, positive eigenvalues, etc.)

This should generalize easily to the complex/Hermitian case.

[dataflow]

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.

[necroforest]

>it's actually impossible to have a non-Hermitian matrix C such that x†Cx > 0

Nice catch, it's been a few years since I had to think about these details.

[hgomersall]

I spent ages looking for a proof of the complex cases during my PhD. Most proofs of CG begin "assuming a real, positive definite matrix".

[dataflow]

Yeah, the complex case seems to get ignored in many areas. I'm guessing it's not particularly useful for many applications.