Is $A$ positive definite iff $\dfrac{A + A^H}{2}$ positive definite? Here define a complex matrix $A\in\Bbb C^{n\times n}$ as being positive definite if $x^HAx>0$ for any $x \neq 0$, where $(\cdot)^H$ denotes conjugate transpose. Note we do not restrict $A$ as a Hermitian matrix. My question is,

Under above definition, is $A$ positive definite iff $\dfrac{A + A^H}{2}$ positive definite? 

My attempt:
If $A$ is positive definite, then $x^H Ax>0$ and it is real, then $x^H Ax=(x^H Ax)^H=x^H A^H x$, and clearly $x^H (\frac{A^H+A}{2})x>0$.
I have trouble with the converse. If $\frac{A^H+A}{2}$ is positive definite, then $x^H Ax+x^H A^H x>0$ and is real, then $x^H A^H x= \overline{x^H Ax}=x^T \bar A \bar x$ for any $x\in \Bbb C^n$ (yes I realized this equality is wrong, thanks to comment by @KittyL). By taking $x = e_1,...,e_n$, where $e_i$ is a vector with the $i$th component being $1$ and all other components being $0$, we have $A^H = \bar A$ or $A = A^T$. I get stuck here. How do we derive $x^HAx>0$ for any $x \neq 0$ from $A=A^T$?
Thanks!
PS: it is possible that the claim is false. If so, please help provide some counter-example.
 A: First of all, to repeat the Wikipedia article on positive-definitive matrix:

In linear algebra, a symmetric $n\times n$ real matrix $M$ is said to be positive definite if the scalar ${\displaystyle z^{\mathrm {T} }Mz}$  is positive for every non-zero column vector ${\displaystyle z}$  of  $n$ real numbers. Here ${\displaystyle z^{\mathrm {T} }}$ denotes the transpose of ${\displaystyle z}$.
More generally, an $n × n$ Hermitian matrix ${\displaystyle M}$  is said to be positive definite if the scalar ${\displaystyle z^{*}Mz}$   is real and positive for all non-zero column vectors $z$ of $n$ complex numbers. Here ${\displaystyle z^{*}}$  denotes the conjugate transpose of $z$.

Your definition is an illed-defined one. For instance when $n=1$, $A=[i]$ and  $x=1$, $x^TAx$ and $0$ are not comparable. In genearl, then one can show that if $A$ is an $n\times n$ complex matrix which is not Hermitian, there exists $z\in\mathbb{C}^n$ such that $z^*Az$ is not real thus it does not make sense to talk about the condition $z^*Az>0$.


Under above definition, is $A$ positive definite iff $\frac{A + A^H}{2}$ positive definite?

Assume now you are using the "correct" version of definition. The condition on $A$ would be trivial: in order to talk about "positive-definiteness", $A$ must be Hermitian. On the other hand, once $A=A^H$, the statement is trivial.
A: Notice that every matrix A can be written as $A=H_1+iH_2$, where $H_1,H_2$ are Hermitian. 
Thus, for every $x$ and $j=1,2$, $x^HH_jx\in\mathbb{R}$ (why?).
Notice that $0<x^HAx=x^HH_1x+i(x^HH_2x)$, for every $x$, implies $x^HH_2x=0$, for every $x$, and $H_2=0$ (why?). 
So $A=H_1=\dfrac{A+A^H}{2}$.
A: I¡-i'd say it does not hold the other way. Lets say the matrix is diagonalizable $A=S D S^{-1}$. Then:
$x^H (A^H + A) x = x^H S (D^H + D) S^{-1} x > 0$
This will hold as long as $(D^H + D)>0$ but this does not precisely mean that $D>0$. An counter example matrix would be:
$$
A = [\begin{matrix}
        1+i & 0 \\
        0 & 1  \\
        \end{matrix}]
$$
Here $(A^H+A)$ is positive definite, but A is not. For $x=[1 \ \ 0]^T$,  $x^H (A^H + A) x = 1+i$ is not real and therefore the standard 'greater than' expression does not apply.
