Does this type of matrix have orthogonal eigenvectors? In this paper, it is claimed that the following matrix has orthogonal eigenvectors:
$$M=\left[\begin{matrix} I & \Delta A \\ -\Delta A^T & I + \Delta^2 A^T A\end{matrix}\right]$$
where $A$ is a real matrix, $I$ is the identity matrix, and $\Delta$ is a real number.
$M$ is neither Hermitian nor anti-Hermitian, either of which would imply having orthogonal eigenvectors.
Does $M$ have orthogonal eigenvectors? How can this be proven?
 A: My first thought was that you copied the lower left block incorrectly, but no, there is actually a negative sign in that paper. Without that negative sign, then the claim would clearly hold (symmetric matrix). 
Counter example: $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $\Delta=1$. Two of the eigenvalues are $\lambda_\pm = \frac{1}{2} ( 3 \pm i \sqrt{3})$ with corresponding eigenvectors $v_\pm = \begin{bmatrix} \frac{1}{2}(1\mp  i \sqrt{3}) & 0 & 0 & 1 \end{bmatrix}^T$, which are not orthogonal. 
Most other matrices I tried for $A$ seemed to fail the claim as well. Unless there's some other condition floating around in that paper beyond the stated assumptions, I think it's an error. 
A: They have actually calculated the eigenvectors explicitly in (41) as (skipping the inverted breve accenting)
$$
\begin{bmatrix}h_i'\\\alpha e_i'\end{bmatrix},\quad\alpha\in\Bbb C,\tag{41}
$$
where
$$
\begin{bmatrix}h_i'\\e_i'\end{bmatrix}\tag{37}
$$
are eigenvectors to the skew-symmetric matrix in (36), thus, orthogonal. Together with $e_i'\bot e_j'$ in (19) it makes orthogonality for (41) as well. I believe it is a simple physical fact (that I do not understand).
