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, 2012 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, 2012 at 15:43

1 Answer 1


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? $\endgroup$ Nov 27, 2012 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .