# A special eigenvalue problem

For my research I need to solve a generalised eigenvalue problem $Ax=\lambda B x$, where $A$, $B$ are general matrices, and selectively find only eigen-pairs $\lambda, x$ such that $\lambda\in \mathbb{R}$ and $x^T$ is also left eigenvector without solving the full problem. Does this problem has a name? I would appreciate some references. I am interested in iterative algorithms for its solution, such as conjugate gradient. However, people typically put additional constraints that $A$ and $B$ are symmetric and $B$ is positive definite. In the present case it is known that $A$ and $B$ are not symmetric.

• Do you have any theoretical property on $A$ and $B$ that guarantees that there are nontrivial solution? Because I am afraid that the general case is that there is none. – Federico Poloni Aug 22 '17 at 8:16
• I googled a bit and it seems that there are some results on the eigenvalues of a product (in your case: $AB^{-1}$), especially when the matrices are symmetric and positive definite. You might be able to find an algorithm that allows for the computation of these eigenvalues without actually computing the product or the inverse if you look in this direction. Unfortunately, I'm not familiar enough with these topics to help you more. – Dirk Aug 22 '17 at 8:42
• @DirkLiebhold I think that the main difficulty here is knowing which eigenpairs to look for. – Federico Poloni Aug 22 '17 at 9:05
• @IgorKhavkine This could be a good idea. Assuming a good starting guess is known, what would be the procedure for iterative refinement ? – yarchik Aug 22 '17 at 11:45
• @yarchik It makes sense to worry because the critical point of $x^TAx/x^TBx$ are solutions of the equation $(A+A^T)x=\lambda (B+B^T)x$. – Surb Aug 22 '17 at 14:22