Recently I have encountered the following problem: I have a $2\times2$ digonalizable matrix of the following form: $A = \begin{pmatrix} a_{11} & a_{12} \\ -a_{12} & a_{22} \end{pmatrix}$, where $a_{k,l} \in \mathbb{C}$. Clearly, this matrix is not some random $2 \times 2$ matrix with complex entries, but has a certain symmetry $A_{12} = - A_{21}$. It is almost a complex skew-symmetric, except that diagonal entries are non-zero. Moreover, the matrix $A$ has left eigenvectors being not equal or somehow simply related to the right eigenvectors $\Big( \: \mathbf{v}^{(L,k)} \ne \left(\mathbf{v}^{(R,k)}\right)^T$, or $\mathbf{v}^{(L,k)} \ne \left(\mathbf{v}^{(R,k)}\right)^\dagger \Big)$, $k$ labels the eigenvectors, $T$ is transpose $^\dagger$ is Hermitian conjugated. So, left eigenvectors can only be constructed if one forms a transformation matrix that consists of right eigenvectors as its columns, inverts it, and then the rows of the result are the left- eigenvectors, namely: $$ S = \begin{pmatrix} \mathbf{v}^{(R,1)}, \mathbf{v}^{(R,2)} \end{pmatrix} , \quad \quad S^{-1} = \begin{pmatrix} \mathbf{v}^{(L,1)} \\ \mathbf{v}^{(R,2)} \end{pmatrix}.$$
Finally, my goal was to somehow alter the matrix $S$ such that right- and left- eigenvectors are somehow nicely related. In this regard, the only degree of freedom I have is normalization of right eigenvectors as they are defined up to a constant, so $S'=\begin{pmatrix} c_1\mathbf{v}^{(R,1)}, c_2\mathbf{v}^{(R,2)} \end{pmatrix}$, where $c_1, c_2 \in \mathbb{C}$. What I found through an observation is that by properly choosing $c_1, c_2$ I can have the following:
$$ S = \begin{pmatrix} v^{(R,1)}_1 & v^{(R,2)}_1 \\ v^{(R,1)}_2 & v^{(R,2)}_2 \end{pmatrix}, \quad S^{-1} = \begin{pmatrix} v^{(L,1)}_1 & v^{(L,2)}_1 \\ v^{(L,1)}_2 & v^{(L,2)}_2 \end{pmatrix} = \begin{pmatrix} v^{(R,1)}_1 & -v^{(R,1)}_2 \\ -v^{(R,1)}_1 & v^{(R,2)}_2 \end{pmatrix},$$ and, probably, this is the best thing I can have for such a matrix.
However, I feel like inventing a bicycle here, mb it is some well-known property of matrices obeying $A_{ij}=-A_{ij}$ for $i \ne j$, and may be there is a more general statement for matrices of arbitrary size with this type of symmetry? Is it something well-known?
P.S. What I know is that for complex symmetric matrices of any size obeying $A_{ij} = A_{ji}$ it is possible to normalize the right eigenvectors such that the left eigenvectors are just transposed of the right ones, namely $\mathbf{v}^{(L,k)} = \left( \mathbf{v}^{(R,k)} \right)^T$ for any $k$. And the normalization constants should be chosen such that $\sum\limits_{l=1}^{N} \left( \mathbf{v}^{(R,k)}_l \right)^2 = 1$ (mind that no complex conjugation!). I found this statement without proof in some article on theoretical physics, and tested it numerically, it works! Gave P.S. just for a reference.