# Are the eigenvalues of the symmetric part of a matrix with eigenvalues - all with positive real part- positive too?

Consider the invertible matrix $$A$$ with all eigenvalues $$\lambda_{i}$$ with positive real part. Then consider its symmetric part: $$A_{s}=\frac{A+A^{T}}{2},$$ with real eigenvalues $$\eta_{i}$$. Are these eigenvalues $$\eta_{i}$$ all positive?

P.S. What I already know is that $$\text{Trace}[A_{s}]=\text{Trace}[A]$$.

No. E.g. the eigenvalues of $$\dfrac12\left(\pmatrix{1&4\\ 0&1}+\pmatrix{1&0\\ 4&1}\right)=\pmatrix{1&2\\ 2&1}$$ are $$3$$ and $$-1$$.
The converse is true, however. That is, if $$A$$ is a real square matrix such that $$\frac{A+A^T}{2}$$ is positive definite, then all eigenvalues of $$A$$ have positive real parts. To prove this, note that if $$u$$ is a unit eigenvector of $$A$$ corresponding to an eigenvalue $$\lambda$$, then $$\Re(\lambda) =\frac{u^\ast(Au)+(u^\ast A^\ast)u}{2} =u^\ast\frac{A+A^\ast}{2}u =u^\ast\frac{A+A^T}{2}u>0.$$
The matrix $$A = \begin{bmatrix}1 & 2 \\ 0&1\end{bmatrix}$$ has only one eigenvalue: $$1$$.
But $$A+A^T = \begin{bmatrix}2 & 2 \\ 2&2\end{bmatrix}$$ has the eigenvalue $$0$$.