# Approximate inverse to a singular diagonal matrix

Suppose that $$A$$ is a square sparse matrix that is diagonal apart from one entry

$$A = \begin{pmatrix} a_{11} & & \\ & \ddots & \\ & & a_{nn} \end{pmatrix}$$ with all $$a_{ii}$$ non-zero apart from one entry. Otherwise, the values are quite large $$a_{ii} \approx 10\rightarrow 1000$$ range.

Can I use a Neumann series to approximate the inverse matrix $$A^{-1}$$? Or, is there any way to approximate the inverse?

• Does "diagonal apart from one entry" mean that there is a non-zero off-diagonal entry that you have otherwise made no mention of? araomis is correct that $A$ will still be singular ($A$ is triangular, and the eigenvalues of a triangular matrix are its diagonal elements, one of which is $0$). But the rest of the discussion in that post applies to a strictly diagonal matrix. – Paul Sinclair Jul 18 at 0:43

My thoughts are the following: First note that strictly speaking there is no such inverse because your matrix is singular. But let $$a_{i,i}$$ be the zero entry on the diagonal. Then set $$a_{i,i} = \epsilon$$. If you think about the inverse of the matrix you get, it'll be the matrix with the entries $$b_{j,j} = \frac{1}{a_{j,j}}$$ on the diagonal. Now as $$\epsilon \rightarrow 0$$ all entries stay constant except $$b_{i,i} \rightarrow \infty$$. So in general the larger you choose $$b_{i, i}$$, the better your approximation will be.