Suppose $\mathbf{A}\in \mathbb{R}^{n\times n}$ is non-singular and its inverse $\mathbf{A}^{-1}$ is known. We would like to compute $\mathbf{B}^{-1}$ where we assume that $\mathbf{B} = \mathbf{A}+\mathbf{I}$ is also non-singular. Directly computing the inverse of $\mathbf{B}$ requires the order of $O(n^3)$. My question is that is there any way of computing $\mathbf{B}^{-1}$ with reduced complexity, say $O(n^2)$?
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How can you be sure $B$ is non-singular? – Bernard Jul 07 '17 at 17:56
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Will the Sherman Morrison formula help? https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula – user1952500 Jul 07 '17 at 18:00
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@user1952500 not a good idea; the matrices both have full rank. Perhaps the Woodbury identity could work, though – Ben Grossmann Jul 07 '17 at 18:04
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If $|\mathbf A|$ is sufficiently small, the inverse can be computed via geometric series, which has linear convergence. – Ben Grossmann Jul 07 '17 at 18:05
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@Omnomnomnom thanks for the comment. – user1952500 Jul 07 '17 at 19:22
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@Bernard Suppose $\mathbf{B}$ is also non-singular. – GOAT Messi Jul 08 '17 at 19:44
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Check out this thread – user3658307 Jul 31 '17 at 17:59