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## Re: solving Ax=b

 From: A S Hodel Subject: Re: solving Ax=b Date: Sat, 4 Jan 2003 07:04:21 -0600

For your problem I suspect that "easy" is a relative term (like NASA's goal of "low-cost" access to space).


There are numerous iterative methods for large, sparse A x = b problems. Most of these will require work on your part. See Golub and Van Loan, "Matrix Computations," Johns-Hopkins University press, for an introduction to some of them.


As a starting point, you may wish to look at the GMRES method by Youcef Saad (I think this is published in SIAM J. Sci. Stat. Comput. 1986). The octave routine krylov may be a useful starting point, but you'll have to modify it to use your f(A,x) instead of A*X. If you can look at other splittings A = M + N besides M= I, N = (A-I), then some of Gene Wachspress's ADI (Alternating Direction Implicit) methods of the 60's may be of use. I haven't kept up with the latest and greatest very large sparse solver methods, so I'd suggest a visit to the last few years of SIAM J. Matrix Analysis and SIAM J. Sci Stat Comput. to see what you can find.

On Saturday, January 4, 2003, at 03:00 AM, Sven Khatri wrote:


Hi All,

I'm looking to solve a problem that is equivalent to solving for x in
the system of linear equations:
Ax = b
where A \in R^{n \times n}, x \in R^n, b \in R^n. But the catch is
that n is too large to explicitly construct A (even in Matlab's sparse
implementation) but I can construct a function f so that f(x) = Ax. Is

there an easy way (within octave) to solve for x? I can solve the problem
by introducing \tilde{A} = I - A, so that the solution to
x = \tilde{A}x + b
is a solution to the above problem and \tilde{A} is a contraction and
solving the problem iteratively but this computation is VERY slow.

thanx...
Sven

PS I hope you all are comfortable with the above latex notation

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