A question about non-symmetric real matrix Let $A$ be a non-symmetric $n\times n$ real matrix. Assume that all eigenvalues of $A$ have positive real parts. Could you please show some simple conditions such that $A+A^T$ is positive definite? Thank you so much.
 A: Let $\lambda$ be an eigenvector for $A$ with eigenvector $v$, then $A v = \lambda v$ and $v^{\dagger}A^{\dagger} = \overline{\lambda} v^{\dagger}$ ($\cdot^{\dagger}$ denotes as conjugate transpose, notice that for matrix $A$ it holds $A^T = A^{\dagger}$, because we know that the entries of matrix $A$ are real). Then, by multiplying the first equation by $v^{\dagger}$ from the left and the second equation by $v$ from the right we obtain: 
$$
v^{\dagger} A v = \lambda v^{\dagger} v, \\
v^{\dagger} A^{\dagger} v = \overline{\lambda} v^{\dagger} v.
$$
Hence, $v^{\dagger} (A + A^{\dagger}) v = (\lambda + \overline{\lambda}) v^{\dagger} v = 2\Re(\lambda) \cdot |v| > 0$, if $v \neq 0$. Notice that we worked out this for any pair of eigenvalue-eigenvector $(\lambda, v)$ and since any other vector can be represented by a linear combination of eigenvectors, then this argument remains valid for all $x$, i.e. $x^{\dagger}(A + A^T)x > 0$ for all $x \neq 0$.
So, matrix $A + A^T$ is indeed positive definite without any conditions. 
EDIT:
As it was noticed by multiple users the conclusion of my answer isn't correct (see comments) and I don't see an obvious solution how to fix this in order to claim that $x^{\dagger}(A + A^T)x > 0$ for all $x \neq 0$. Thanks to @user1551.
A: As the OP writes, that follows is a class of $A$ s.t. $A+A^T>0$.
Choose $A=[a_{i,j}]$ s.t. for every $i$, 
$2a_{i,i}>\sum_{j\not= i}(|a_{i,j}+a_{j,i}|)$;
