Eigenvalue perturbation with complex-symmetric (non-Hermitian) matrices The Rayleigh-Schrodinger method to find the perturbation to an eigenvalue of a Hermitian matrix is standard and well-known. How to find the perturbation to a non-degenerate eigenvalue if the unperturbed and the perturbated matrices are complex symmetric (hence non-Hermitian), instead of Hermitian?
In other words, if you have non-degenerate eigenvalues and eigenvectors of a complex-symmetric matrix $H_0$, how would you find the eigenvalues of the matrix $H_0 + V$, where $V$ is a "small" perturbation to $H_0$ and is also complex-symmetric?
 A: I will use Dirac's bra-ket notation but in a different sense. $\langle a|$ stands for the transpose of $|a \rangle$, rather than its conjugate transpose (the standard usual sense). Hence, $\langle a|b \rangle  \equiv  a^T b$ and $ \langle a|M|b \rangle \equiv a^T M b$.
Now, the unperturbed eigenvalue equation is (following the presentation style of Sakurai's Modern Quantum Mechanics section 5.1).
$ H_0 |n_0  \rangle =   E^0 | n_0 \rangle   $,
whereas the perturbed equation is
$ (E^0 - H_0) |n \rangle  =  (V - \Delta) | n  \rangle $.
Here $H_0, E^0$ and $|n_0  \rangle$ are the unperturbed complex-symmetric (CS) matrix, unperturbed eigenvalue, and the unperturbed eigenvector, respectively, whereas $V, \Delta$ and $ | n \rangle$ are the CS perturbation matrix, change in the eigenvalue and the perturbed eigenvector.
The above equation gives us
$ \langle n_0 |(E^0 - H_0) |n \rangle  =  \langle n_0 | (V - \Delta) | n  \rangle $.
Since $H_0$ is a CS matrix we have for any general vectors  $|a \rangle$ and $|b \rangle$ with complex entries, that $   \langle a | H_0 | b \rangle = \langle b | H_0 | a \rangle  $. This can be a small exercise problem to prove. $   \langle a | H_0 | b \rangle = \langle b | H_0 | a \rangle  $ makes it possible to write the above equation as
$   \langle n_0 | (V - \Delta) | n \rangle  = \langle n_0 |E^0 |n \rangle - \langle n_0 | H_0 |n \rangle  =  \langle n_0 |E^0 |n \rangle - \langle n | H_0 |n_0 \rangle  =  E^0 ( \langle n_0 |n \rangle - \langle n  |n_0 \rangle)    
 = 0 $ since $ \langle a | b \rangle = \langle b | a \rangle$.
This finally yields
$ \langle n_0 | (V - \Delta) | n \rangle  = 0$
$ \Delta = \frac{\langle n_0 | V  | n \rangle }{  \langle n_0|n \rangle } = \frac{\langle n_0 | V  | n_0 \rangle }{  \langle n_0|n_0 \rangle } $ ,
where we have replaced $ | n \rangle   \rightarrow | n_0 \rangle $ since it makes a difference at a higher order in perturbation than what we are concerned here.
We can finally write the final result as
$ \Delta = \langle \hat{n}_0 | V  | \hat{n}_0 \rangle  $,
with $ | \hat{n}_0 \rangle \equiv \frac{| {n}_0 \rangle}{ \sqrt{ \langle  n_0| n_0  \rangle}} $ being the normalized eigenvector, where the normalization is wrt. a dot product which is based on the transpose rather than the conjugate transpose (see my customized bra-ket definitions at the beginning of this answer). Also, the square root in the above expression may give multiple values since $\langle  n_0| n_0  \rangle$ in general is complex. So, we must use the same value of the square root for  $ | \hat{n}_0 \rangle$ and $ \langle \hat{n}_0 |$ in the above equation $ \Delta = \langle \hat{n}_0 | V  | \hat{n}_0 \rangle  $. In case of confusion, one can also use the other expression for $ \Delta = \frac{\langle n_0 | V  | n_0 \rangle }{  \langle n_0|n_0 \rangle } $.
