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The following is from WiKi:

In linear algebra, a symmetric n × n real matrix M is said to be positive definite if zTMz is positive, for all non-zero column vectors z of n real numbers; where zT denotes the transpose of z. More generally, an n × n complex matrix M is said to be positive definite if z*Mz is real and positive for all non-zero complex vectors z; where z* denotes the conjugate transpose of z. This property implies that M is an Hermitian matrix.

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closed as not a real question by Ryan Budney, Pietro Majer, Emil Jeřábek, Chris Godsil, Andreas Blass Dec 4 '12 at 18:22

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The point is that $z^T M z$ defines a quadratic form. If $M$ is not symmetric it can be replaced with $1/2(M+M^T)$ and with that matrix in place of $M$, exactly the same quadratic form results! check it. – kjetil b halvorsen Dec 4 '12 at 17:43
The partial order for symmetric matrices is defined as $A\ge B$ iff $A-B$ is positive. If you try to define it for all matrices, most properties would be destroyed. Check the proofs. – Pietro Majer Dec 4 '12 at 17:49
Can you provide the link to the Wikipedia page? If it really says that "This property implies that $M$ is an Hermitian matrix", then this is simply wrong and should be corrected. For example, if $M$ is anti-hermitian, then $z^T Mz=0$ for all $z$. – Tobias Fritz Dec 4 '12 at 18:07

In real domain, $M$ is positive definite means $z^{T}Mz>0$ for non zero $z$. It does not means $M$ is symmetric. More generally this means symmetric part of the matrix $(M^{T}+M)/2$ is positive definite.

While in case of complex $z^{H}Mz$ is real for all $z$ implies matrix is hermitian. Combined with $z^{H}Mz>0$, it is positive definite. ( when you say $z^{H}Mz>0$ for all $z$ it is automatically means that $z^{H}Mz$ is real for all $z$ as well.

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