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?

  • $\begingroup$ 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. $\endgroup$ – 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.


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