Say we have a matrix $A = L + \beta^{2} M$, where $\beta$ is a real scalar. The matrices $L$ and $M$ are symmetric positive semi-definite and symmetric positive definite respectively. I am interested in obtaining a low-rank approximation of $A^{-1}$, which can be obtained by solving the GHEP given by $L U = MU\Lambda$, where $U$ is the eigenvector matrix and $\Lambda$ is a diagonal matrix of the corresponding eigenvalues. Using this approach, $A^{-1} = U(\Lambda+\beta^2I)^{-1}U^{\mathrm{T}}$.
Now, consider the case when $\beta=0$. From the above equation, $A^{-1} = U\Lambda^{-1}U^{T}$. However, since $\beta=0$, we have $A=L$. This means that I can also compute $A^{-1}$ by solving the HEP given by $LV= \Omega V$ and $L^{-1} = V\Omega^{-1}V^{T}$.
I have the following questions regarding $A^{-1}$ and $L^{-1}$ for this case when $\beta=0$.
- I calculated $A^{-1}$ and $L^{-1}$ by solving the corresponding eigenvalue problems and computing the first 100 eigenpairs, I see that $A^{-1}x$ and $L^{-1}x$ (for some vector $x$) are quite different. Is this expected? If so why?
- Does $A^{-1}x$ approach $L^{-1}x$ as the number of eigenpairs used to compute $A^{-1}$ is increased?
- Will $A^{-1}x$ ever match $L^{-1}x$ in matrices of finite sizes, if $M\neq I$?
