# why do we need to specify to symmetric matrix when defining real positive definite matrix? [closed]

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 BlassDec 4 '12 at 18:22

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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.