Imagine that you have a complex symmetric matrix $M$ of size $N\times N$ such that $M_{jk}=M_{kj}$.
Now we create a new matrix $M'$ by multiplying every element below the main diagonal of the original matrix $M$ by a fixed real-valued constant $\xi$ so that: $M'_{jk}=\xi M'_{kj}$, where it is assumed that $j>k$. The new matrix $M'$ now is not even normal, so that $M'^{\dagger} M' - M' M'^{\dagger} \ne 0$, and when studying the spectrum of $M'$, one has to use left $v^{(L)}_j$, and right eigenvectors $v^{(R)}_j$. Generally, there is no simple relation between the right, and left eigenvectors for non-normal operators, and one has to simply construct the corresponding transformation matrix to the right-eigenspace from right eigenvectors $S = \left( v^{(R)}_1, \: v^{(R)}_2, \: \ldots \:, v^{(R)}_N \right)$, and then invert it $S^{-1} = \left( v^{(L)}_1, \: v^{(L)}_2, \: \ldots \:, v^{(L)}_N \right)^{T}$, and the corresponding left eigenvectors are the rows of the obtained $S^{-1}$.
However, I noticed that for $M'$ described above there is a relation $v^{(L)}_j = c_j v^{(R)}_j$, where $c_j$ is some complex-valued constant (which is different for different $j$'s).
Of course, all of the above suggests that $M$, and $M'$ are diagonalizable.
My question is: whether is it possible to somehow mathematically derive the constants $c_j$? And do $c_j$ depend solely on the entries of the corresponding right eigenvector $v^{(R)}_j$ or they might also depend on the entries of other right eigenvectors?