Non symmetric matrix and its transpose and positive definiteness Suppose we have a non symmetric Matrix $A$. Let also we know that $A+A^T$ is positive definite. Is it the case that A is also positive definite? If A symmetric then its easy to show that $A$ is positive definite because $A=A^T$. So if it is not the case then can we find a 2x2 matrix, $B$ which is non symmetric and $B+B^T$ is positive definite but B is not positive definite?
I have another question. Suppose $A$ is a non symmetric non positive definite matrix. is it possible its inverse is positive definite?
Thanks.
 A: The answer is no, at least with your definition of positive-definite; we can show this directly in the $2 \times 2$ case:
Let the symmetric part of the matrix be
$$ S = \begin{pmatrix} 2a & b \\ b & 2d \end{pmatrix}, $$
and the antisymmetric part $A$ have top-right element $-c$. It is easy to compute the eigenvalues of $S$ and $S+A$. Since $S$ is symmetric, we expect it to have real eigenvalues, and indeed it does, namely
$$ \lambda_{\pm}=a+d \pm \sqrt{(a-d)^2+b^2}. $$
The square root contains a sum of real squares, so is nonnegative. We also require that
$$ a+d>\sqrt{(a-d)^2+b^2} $$
in order to have $\lambda_-$ positive.
On the other hand, the eigenvalues of $S+A$ are
$$ \mu_{\pm}=a+d \pm \sqrt{(a-d)^2+b^2-c^2}, $$
so we require
$$ (a-d)^2+b^2-c^2>0 $$
for real eigenvalues, which is certainly not necessary, given that we can choose $c$ as large as we like. Hence we can have complex eigenvalues for $S+A$, even if $S$ has positive ones. On the other hand, supposing that $\mu_{\pm} \in \mathbb{R}$, we do automatically get the other condition:
$$ a+d>\sqrt{(a-d)^2+b^2} >\sqrt{(a-d)^2+b^2-c^2}, $$
since $c^2 > 0$ and the square root is increasing.
