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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
Motorola Labs - Paris +33 1 69 35 48 04 (Ph)
Parc Les Algorithmes, Commune de St Aubin +33 6 72 01 06 33 (Mob)
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