# Why do mathematicians use only symmetric matrices when they want positive semi-definite matrices?

Why do mathematicians use only symmetric matrices when they want positive semi/definite matrices?

I mean I haven't seen using non-symmetric positive semi/definite matrices. If non-symmetric positive semi/definite matrices exist can those be always written by a symmetric PSD matrix?

• Can you give a use case for non-symmetric positive-semidefinite matrices?
– Mico
Oct 6 '13 at 13:00
• @Mico Sorry, I don't know. My question is actually can they be? Oct 6 '13 at 13:15
• Here's a problem involving a non-symmetric PSD real matrix: math.stackexchange.com/questions/3855621/…. Oct 7 '20 at 19:12

Edit: here is why in the complex case, positive semidefinite implies hermitian. Actually, the proof implies that in the complex case $A$ is hermitian if and only if $x^*Ax\in\mathbb R$ for all $x$.
Assume $x^*Ax\in\mathbb R$ for all $x$. then $$\mathbb R\ni(y+\alpha x)^*A(y+\alpha x)=y^*Ay+\overline\alpha\,x^*Ay+\alpha\,y^*Ax+|\alpha|^2\,x^*Ax.$$ As this expression is real, it equals its complex conjugate $$y^*Ay+\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y+|\alpha|^2\,x^*Ax.$$ So $$\overline\alpha\,x^*Ay+\alpha\,y^*Ax=\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y.$$
Taking first $\alpha=1$ and then $\alpha=i$, we get $$x^*Ay+y^*Ax=y^*A^*x+x^*A^*y,$$ $$-i\,x^*Ay+i\,y^*Ax=i\,y^*A^*x-i\,x^*A^*y.$$ Multiplying the first equation by $i$ and adding, we get $$2i\,y^*Ax=2i\,y^*A^*x.$$ As this works for any $x,y$, we deduce that $A=A^*$.