# Solving Generalized Eigenvalue Problem perturbatively

Let me formulate the problem to convey my notation. I have a matrix $A$ which is hermitian - and is diagonalisable by a transformation $$U_A A\,\,U_A^{-1} = A_{diag}$$

Now the matrix is changed, using the small parameter $\lambda$. Therefore, $$A \rightarrow A-\lambda B$$ where $B$ can be not Hermitian, in general.

I tried doing this the following way -

The matrix $A-\lambda B$ has to be rotated by $U^\prime = e^{i\lambda \alpha}U_A$ to be diagonalized, $\alpha$ being a generator of rotation - and it makes sense that the "extra" rotation has to be proportional to $\lambda$.

$$A^\prime_{diag} = A_{diag} + C_1\lambda + C_2\lambda^2 + \mathcal{O}(\lambda^3)$$

I wrote this down using the BCH formula and set the linear coefficient, $C_1= 0$ and got the constraint $U_AB\;U_A^{-1} = \left[i\alpha,A_{diag}\right]$.

The final form I obtained for $A^\prime_{diag}$ was $$A^\prime_{diag} = A_{diag} -\frac{1}{2}\left[i\alpha,U_AB\;U_A^{-1}\right]\lambda^2 + \mathcal{O}(\lambda^3)$$

Can anyone tell me if -

1. I am on the right track - and how to proceed from here to solve for $\alpha$ in terms of the known matrices.

2. I need to approach this differently.

3. it would help if $B$ were Hermitian?

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Fun word-perturbatively –  adam W Mar 7 '13 at 17:05
sorry! Is that a non-existent made-up word or do you just find the sound of the word funny? I am a physics student .. and I remember reading about some transformation or other two-three years ago! I loosely remember the principles - and I am trying to reconstruct the method and apply it to my problem! I am thinking of a perturbation in the matrix-space - in the direction of $B$ and the system reacts linearly in $\alpha$. The analogy makes sense to me. –  Debanjan Basu Mar 7 '13 at 17:12
I might start saying it, it is a fun word to me. Your question is difficult for me so I can not really add much else. I can say that to diagonalize the perturbated matrix will require more than just a scalar change in $U_A$ as the eigenvectors will change (unless $B$ shares the same eigenvectors). –  adam W Mar 7 '13 at 17:50
You may also want to tag this physics or mathematical-physics or both, maybe even add the eigenvalues tag. I think it may help gain the proper attention to this question. –  adam W Mar 7 '13 at 18:33
"perturbatively" and "non-perturbatively" are quite standard terms in physics (as you can also ascertain by googling them). –  joriki Mar 7 '13 at 19:24