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?

  • $\begingroup$ Can you give a use case for non-symmetric positive-semidefinite matrices? $\endgroup$
    – Mico
    Oct 6, 2013 at 13:00
  • $\begingroup$ @Mico Sorry, I don't know. My question is actually can they be? $\endgroup$
    – triomphe
    Oct 6, 2013 at 13:15
  • $\begingroup$ Here's a problem involving a non-symmetric PSD real matrix: math.stackexchange.com/questions/3855621/…. $\endgroup$
    – dv_bn
    Oct 7, 2020 at 19:12

1 Answer 1


First, one can argue that non-symmetric positive definite matrices are pathological, in the sense that when you move to the complex case all positive definite matrices are hermitian.

For a non-symmetric positive definite matrix you can say little more than the fact that it has positive eigenvalues. You don't have many of the nice properties that symmetry adds. For instance, without symmetry you don't even have that the singular values agree with the eigenvalues, nor diagonalizability.

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 $A=A^*$.

  • $\begingroup$ Thank you. Could you please explain or provide a reference to a proof where "complex case all positive definite matrices are Hermitian"? Why can't there be a non Hermitian SPD in complex? $\endgroup$
    – triomphe
    Oct 6, 2013 at 13:30
  • $\begingroup$ See the edit, please. $\endgroup$ Oct 6, 2013 at 16:55

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