I wish to invert $\mathbf{A}=(\mathbf{I} - \mathbf{X})$, where $\mathbf{I}$ is the an identity $N\times N$ matrix and $\mathbf{X}$ is an $N \times N$ non-invertible matrix. One way to do it is by using the Newman series expansion, but in my application, it is not converging to the exact solution even when the order of the approximation is very high. I have also tried to do eigen-decomposition, i.e. letting $\mathbf{A}=\mathbf{I} - \mathbf{Q}\mathbf{D}\mathbf{Q}^{-1}=\mathbf{Q}(\mathbf{I} - \mathbf{D})\mathbf{Q}^{-1}$, then after all $\mathbf{X}$ is singular. Is there any way to invert $\mathbf{A}$ exactly without having to invert the whole $(\mathbf{I} - \mathbf{X})$ or $\mathbf{X}$?
Thanks,
_L
I can give you a few methods of calculating the inverse without directly calculating the inverse of $A$, but I can't really speak to their efficacy.
Singular Value or Eigenvalue Decomposition:
SVD tells us that $A=USV^T$ where $U,V$ are orthogonal and $S$ is diagonal. It follows that $A^{-1}=VS^{-1}U^T$. Note that the elements of $S^{-1}$ are simply the reciprocal of the elements of $S$. Eigenvalue decomposition is similar but requires a full set of eigenvalues.
Series Expansion
If we treat $I-X$ as a matrix function, we can expand it's inverse into a taylor series. Note that this requires that $|\lambda|<1$ for all eigenvalues of $I-X$ for convergence:
$$(I-X)^{-1}=I+X+X^2+X^3+...$$
This is an approximation however, and has a pretty strict requirement on the eigenvalues.
Don't Invert it
Many applications that require matrix inversion can be solved by other means such as LU or SVD decomposition. It really depends on why you need to invert this matrix.