Important to note that this method only works on Hermitian (usually AKA symmetric) and positive-definite matrices, both of which are often pretty big qualifiers.
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 nu…
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, = - => = 0. So for any matrix where > 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.