TITLE: Positive definite matrix must be Hermitian
QUESTION [16 upvotes]: Is there a simple way to show that a positive definite matrix must be Hermitian?
I feel there is a long drawn out proof of this to be had by taking unit vectors and applying the positive definiteness property, and brute forcing it.
But is there some simple clever proof why a positive definite matrix is necessarily Hermitian?

REPLY [5 votes]: You don't even need positive definiteness (or semi,negative etc definitiveness for that matter)
Only thing you need is that $x^*Ax \in R \; \forall x \in C$ which is ofcourse true when we compare this value with $0$ while defining such matrices.
So let $x^*A x\in R \; \forall x\in C$ (and NOT just $R$)
I will show $A$ is hermitian
$x^*Ax= \langle Ax,x\rangle=\langle x,A^*x\rangle = \overline{\langle A^*x,x\rangle} = \langle A^*x,x \rangle  \rightarrow \langle (A-A^*)x,x\rangle = 0 \; \forall x\in C$
Claim: if $\langle Bx,x\rangle = 0 \; \forall x\in C \rightarrow B=0$
Proof: For arbitrary $y\in C$, $\; \langle B(x+ky), x+ky\rangle = \bar{k}\langle Bx,y \rangle + k\langle By,x \rangle$
Now set $k=1$ and $k=\iota$ to get two equations, solve them to get $\langle Bx,y\rangle =0 \; \forall y\in C \rightarrow Bx= 0 \; \forall x\in C \rightarrow B=0$