Sufficient condition for a matrix to be hermitian Let $A\in \mathcal{M_{n\times n}}(\mathbb{C})$ be a matrix.
Does $x^*Ax>0$ for all $x\in \mathbb{C}^{n\times 1}$ such that $x\neq 0_{\mathbb{C}^{n\times 1}}$ imply that $A$ is hermitian?
Please provide a proof or a counter-example accordingly.
Are positive-definite matrices defined for all matrices or just for hermitian ones?
 A: If $x^* A x$ is real, that says $x^* A x = x^* A^* x$.
Now the polarization identity
$$ x^* A y = \frac{1}{4} \left(  (x + y)^* A (x + y) 
- (x - y)^* A (x - y) - i (x + i y)^* A (x + i y) + i (x - i y)^* A (x - i y)\right)$$
determines $x^* A y$ in terms of $z^* A z$ for some vectors $z$.
Thus if $x^* A x = x^* A^* x$ for all $x$, we also have $x^* A y = x^* A^* y$ for all $x$ and $y$, and then $A = A^*$, i.e. $A$ is hermitian.
A: Yes. Actually, $x^*Ax\in\mathbb R$ for all $x\in\mathbb C^n$ is enough to force $A$ to be hermitian. 
Given $\alpha\in\mathbb C$, we have
$$
(x+\alpha y)^*A(x+\alpha y)=x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay.
$$
As the term on the left is real, we have that the right hand side equals its conjugate:
$$
x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay=x^*A^*x+|\alpha|^2y^*A^*y+\alpha x^*A^*y+\overline\alpha y^*A^*x.
$$
As $x^*Ax, y^*Ay\in\mathbb R$, we have $x^*A^*x=x^*Ax$, $y^*A^*y=y^*Ay$. Then
$$
\overline\alpha y^*Ax+\alpha x^*Ay=\alpha x^*A^*y+\overline\alpha y^*A^*x.
$$
Taking $\alpha=1$, we get
$$\tag{1}
y^*Ax+x^*Ay=x^*A^*y+y^*A^*x.
$$
Taking $\alpha=i$, we get
$$\tag{2}
-y^*Ax+x^*Ay=x^*A^*y-y^*A^*x.
$$
Subtracting $(2)$ from $(1)$, we get
$$
y^*Ax=y^*A^*x.
$$
As $x,y$ we arbitrary, we conclude that $A=A^*$.
Regarding your second question, there is no agreement. Some require positive-definite matrices to be hermitian, while others simply ask for $x^*Ax>0$. The two things are the same in the complex case, but not in the real one. 
