Positive definite matrix must be Hermitian 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?
 A: As Marvis says in the comments, the problem reduces to showing that if $V$ is a finite-dimensional complex inner product space and $A : V \to V$ an operator such that $\langle v, Av \rangle = 0$ for all $v$, then $A = 0$. Letting $v$ run across all eigenvectors of $A$, the hypothesis implies that $A$ has all eigenvalues $0$, hence is nilpotent. Suppose by contradiction that $A \neq 0$, hence there exists a vector $v_0$ such that $v_1 = A v_0$ is nonzero. We may assume WLOG that $A v_1 = 0$, hence
$$\langle v_0 + v_1, A(v_0 + v_1) \rangle = \langle v_1, v_1 \rangle \neq 0$$
which is a contradiction. Hence $A = 0$. 
Note that the corresponding assertion for real inner product spaces is false; the condition $\langle v, Av \rangle = 0$ is satisfied by all skew-symmetric matrices. 
A: 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 $\langle Bx,y\rangle =0 \; \forall y\in C \rightarrow Bx= 0 \; \forall x\in C \rightarrow B=0$
