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Hi, everyone.

What I am struggling is the following problem. I have a linear matrix equation $Ax=b$, where $A$ is a known $n \times n$ large sparse real matrix, $x$ and $b$ are known $n \times 1$ vectors. Now, one entry of $A$ has changed into $a$ and we denote by this matrix $A'$. Since we make $b$ unchanged, the updated $A$ will cause the $x$ in the original linear eqaution accordingly changed to $x'$ such that $A'x'=b$. My goal is to find this new $x'$. A naive way is to re-solve $A'x'=b$. But since $A'$ is slightly different from $Ax=b$, is there any incremental way to fast solve $x'$ in $A'x'=b$ by taking advantage of the original equation $Ax=b$? Thank you very much for any of your kind suggestion!

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4 Answers 4

Another method to update the solution is using the Sherman-Morrison formula: http://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula in your case, $u$ and $v$ are canonical basis vectors.

So basically you have to solve two linear systems with $A$ and then you can update the solution for all possible values of $A$ with little work. Solve $2n$ linear systems, and you can update as many times as you want every entry of $A$ (only one at a time though).

Not sure that this is really your best option though --- all depends on how many "modified systems" you have to solve with the same starting matrix $A$. We need more information from you to decide this.

[By the way, as already pointed out, you'd better use a linear system solver which is suitable for sparse matrices: sparse LU or iterative methods.]

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An iterative scheme may do the trick. I would suggest looking into algorithms such as GMRES. Since you have a large, sparse matrix, there is a good chance you already have your matrix in a format that can be accepted by an iterative solver.

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As Fumiyo Eda already mentioned, you can use an iterative method such as GMRES to resolve the system after the change to $A$.

If you want to use direct LU factorization rather than an iterative method, look into "rank one update" techniques for the LU factorization.

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You can update p columns of de A Matrix with Woodbury matrix identity http://en.wikipedia.org/wiki/Woodbury_matrix_identity and solve an pxp system.


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Well, that is basically the same as @Federico already proposed (although your link points to something more general, and in fact, more general than needed here). –  Dirk Sep 10 '13 at 7:08

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