I wanted to ask why is positive definite semi-ordering is well defined only for Hermitian matrices (or symmetric matrices if restricted to the reals)?

I saw an extension to the definition of positive definite matrices to the asymmetrical case. Namely, we say that a real (unnecessary symmetric) matrix $M$ is positive definite in the wide sense if and only if $z^TMz>0$ for all non-zero vectors $z$. This is equivalent to that the symmetrical part of $M$, which is $\frac{M+M^T}{2}$, is positive definite in the narrow sense.

Why is not possible to define a semi-ordering using this extended definition without considering self-adjoint matrices?

Thank yo uall in advance, Ziv Goldfeld

  • 2
    $\begingroup$ Are you sure you mean a "semi-ordering"? I don't think the second condition in the definition of a semiorder is fulfilled. Perhaps you mean a partial order? $\endgroup$ – joriki Nov 26 '12 at 15:33
  • $\begingroup$ I guess it's partial order. How do you compare [0,1;0,0] and [0,0;1,0]? $\endgroup$ – Hui Yu Nov 26 '12 at 15:43

You certainly can say that $A\leq B$ is $B-A$ is positive semi-definite. But to see why people not pay much attention to this, consider what you get when $A,B$ are complex numbers: you would be defining $a+ib\leq c+id$ if $b=d$ and $a\leq c$. For instance, $$ 1+2i\leq 3+2i. $$ Not a lot of information can be gathered from this, as you are basically considering the order of the reals for the real part, and fixing the imaginary part.

  • $\begingroup$ I see, but how does the self-adjoint case solves this problem you presented for the non self-adjoint case? $\endgroup$ – Ziv Goldfeld Nov 27 '12 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.