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Re: matrix inversion
From: |
David Bateman |
Subject: |
Re: matrix inversion |
Date: |
Wed, 22 Nov 2006 17:31:51 +0100 |
User-agent: |
Thunderbird 1.5.0.7 (X11/20060921) |
Gorazd Brumen wrote:
> Hello,
>
> I have a square matrix that is composed of N^2 block matrices, where
> every block matrix is relatively large t x t diagonal matrix (both
> N and t large), i.e.
>
> A = [ A_{11}, A_{12} , ... A_{1N}; A_{21} ... A_{2N} ...],
> where every A_{ij} is a diagonal matrix of size txt.
>
> My question is the following: How many operations would
> spinv need to do this inversion?
>
>
> Thanks and regards,
> Gorazd
>
>
Ok, just for the hell of it I ran
N=5;
t=5;
dt = speye(t);
A = repmat(dt,N,N);
[p,q,r,s] = dmperm(A);
spy(A(p,q))
and I saw that the block triangular factorization of this beast is in
fact N t-by-t full matrices. So yes a BTF technique really does make
sense here.. I believe, that something like
[p,q,r,s] = dmperm(A);
b = speye(rows(A))(p,:);
A = A (p,q);
x = zeros(size(A));
for k = r-1:-1:1
# Get k-th block
k1 = r(k);
k2 = r(k+1) - 1;
# Solve the system
x(k1:k2,:) = A(k1:k2,K1:k2) \ b(k1:k2,:);
# off-diagonal block back subsitution
b (1:k1-1,:) = b (1:k1-1,:) - A (1:k1-1, k1:k2) * x (k1:k2,:) ;
endfor
x(q,:) = x;
might be an efficient manner to solve this type of problem (at least in
theory).
D.
--
David Bateman address@hidden
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