I encountered a few matrices like the ones in the following:
$A_1=\begin{bmatrix}1 & -1 & 0\\0 & 1 & -1\\-1 & 0 & 1\end{bmatrix}$, $A_2=\begin{bmatrix}0.84 & -0.62 & -0.22\\-0.22 & 0.84 & -0.62\\-0.62 & -0.22 & 0.84\end{bmatrix}$ which respectively have eigenvalues $\lambda(A_1)=0,~1.5\pm{j*0.866}$ and $\lambda(A_2)=0,1.26\pm{j*0.3464}$.
What surprises me is $\lambda(A_1+A^T_1)=0,3,3$ and $\lambda(A_2+A^T_2)=0,2.52,2.52$, i.e. the eigenvalues of the resultant matrix are sum of corresponding complex and complex conjugate eigenvalues. Is this associated to any property of matrices which make the resultant eigenvalues of $(A_i+A^T_i)$ twice the real part of eigenvalues of $A_i$?
I have one more matrix to show similar situation.
$V=\begin{bmatrix}-0.5773 & -0.5773 & 0.5773\\-0.5773 & -0.288675 & -0.288675\\-0.5773 & -0.288675 & -0.288675\end{bmatrix}$, which gives
$\lambda(V)=0,-0.5773\pm{j*0.8164}$ and $\lambda(V+V^T)=0,-1.1547,-1.1547$.
My approach: any matrix $A=\dfrac{1}{2}(A+A^T)+\dfrac{1}{2}(A-A^T)$ and for a nonzero vector $x$, we get $x^T{A}x=\dfrac{1}{2}x^T(A+A^T)x$ since $x^T(A-A^T)x=0$ due to skew-symmetricity of $A-A^T$.
Therefore $x^T(A+A^T)x=2x^T{A}x=2x^T{P}DP^{-1}x$, where $A$ is decomposed as $A=PDP^{-1}$ with $D$ being a diagonal matrix with all eigenvalues of $A$ along its diagonal and $P$ is a unitary matrix with complex conjugate transpose denoted by $P^*=P^{-1}$. [The eigenvectors of $A_{1},A_2,V$ are independent and constitute a basis for $\mathbb{C}^3$].
Since $2x^T{PDP^{-1}}x\leq{2}x^T\text{tr}(PDP^{-1}){x}=2x^T\text{tr}(D){x}=4*\text{real}(\lambda)\|x\|^2$. But this does not help me prove the claim. Ant hint or references will be greatly appreciated.