Do the complex conjugate eigenvalues add up in pairs for the addition of a matrix and its transpose?

Question

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.

Answer 1

This property holds here only because all of your examples share very special properties:

1. $A$ is not invertible
2. $A+A^T$ is still not invertible (not automatic, see $A=\left(\matrix{0&0&0\\1&0&0\\1&1&0\\}\right)$)
3. Besides the eigenvalue $0$, $A+A^T$ has only one eigenvalue, say $\lambda,$ of multiplicity $2$.

Whenever all this holds, then since $2\lambda=\operatorname{Tr}(A+A^T)=2\operatorname{Tr}(A)$, you see that $\lambda$ is indeed the sum of the complex eigenvalues of $A$.

Remark. Any real symmetric matrix has real eigenvalues. The fact that the imaginary part disappears is nothing surprising.

Remark 2. There are matrices that satisfy 1. and 2. above, that have non-real eigenvalues, but which do not satisfy 3. (and hence your conclusion fails for these matrices). For instance, Take $$A=\left(\matrix{0&0&0\\0&0&-2\\0&1&0\\}\right)$$ Conditions 1. and 2. are clear. The eigenvalues of $A$ are $0$ and $\pm i\sqrt2$. The eigenvalues of $A+A^T$ are $0$ and $\pm 1$.

Remark 3. This does not mean that you don't have discovered something. But you would need to say a bit more about how you created these matrices so that someone can derive a general proof of what is going on in your specific settings.

Answer 2

Based on the given eigenvalues, there seems to be a typo in $V$: the sign of the $(1,2)$-th entry should be positive instead of negative, i.e. $$ V=\frac1{2\sqrt{3}}\pmatrix{ -2& 2& 2\\ -2&-1&-1\\ -2&-1&-1}. $$ It can be easily verified that $A_1,A_2$ and $V$ are normal matrices. Since normal matrices are unitarily diagonalisable, the said observation follows immediately because you are basically dealing with the scalar case.