Are eigenvalues of $A$ all with positive real parts if and only if $x^TAx>0$ for any $x$? $A$ is non symmetric. If this is true, if $B=-B^T$, then if the eigenvalue of $A$ are with positive real parts, so will be the eigenvalue of $(A+B)$.
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Does this answer your question? If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts? – user1551 Jul 22 '20 at 05:47
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To borrow Erick Wong's answer, if $x^TAx>0$ for all real vectors $x$, then $A+A^T$ is positive definite. Now let $(\lambda,x)$ be an eigenpair of $A$ and $\langle\cdot,\cdot\rangle$ be the standard inner product on $\mathbb C^n$. Then \begin{aligned} \lambda\|x\|^2&=\langle Ax,x\rangle,\\ \overline{\lambda}\|x\|^2&=\overline{\langle Ax,x\rangle}=\overline{\langle x,A^Tx\rangle}=\langle A^Tx,x\rangle. \end{aligned} Hence $2\Re(\lambda)\|x\|^2=\langle (A+A^T)x,x\rangle$. In turn, $\Re(\lambda)>0$ because $A+A^T$ is positive definite.
The converse is false, as shown in Erick Wong's answer (in which $A$ is diagonalisable over $\mathbb R$) or the other answer here (in which $A$ is not diagonalisable over $\mathbb R$).
user1551
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If $x^TAx$ is positive, then the eigenvalues are positive (irrespective of the matrix being symmetric or not), but the converse is not true for non-symmetric matrices. (The converse is true for symmetric matrices.)
For instance, consider the Jordan matrix $$A = \begin{bmatrix}1 & -4\\0 &1 \end{bmatrix}$$ The eigenvalues are clearly positive (since the eigenvalues are both $1$) but the matrix is not positive definite.
user141421
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