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<TITLE>RE : [scilab-Users] inversion of a huge block diagonal matrix</TITLE>
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<P><FONT SIZE=2>>Stupid question first: do you have tested to store your big matrix in a<BR>
>sparse matrix ?<BR>
>stacksize('max')<BR>
>B = spzeros(1e6,1e6);<BR>
>// Store the matrix<BR>
>Binv = inv(B);<BR>
<BR>
<BR>
<BR>
Yes, I have already tested this solution<BR>
But Binv = inv(B) is time consuming too when the size of the matrix is high<BR>
I guess that this way does not take advantage of the fact that the matrix is block diagonal<BR>
<BR>
<BR>
Guylaine<BR>
<BR>
<BR>
<BR>
-------- Message d'origine--------<BR>
De: Collette yann [<A HREF="mailto:yann.collette@scilab.org">mailto:yann.collette@scilab.org</A>]<BR>
Date: mar. 16/03/2010 15:41<BR>
À: users@lists.scilab.org<BR>
Objet : Re: [scilab-Users] inversion of a huge block diagonal matrix<BR>
<BR>
Collewet Guylaine a écrit :<BR>
><BR>
> Hello,<BR>
><BR>
> I would like to compute the inverse of a huge block diagonal matrix<BR>
> using an efficient way.<BR>
><BR>
> Each block is a square matrix with NxN dimension and the block<BR>
> diagonal matrix dimension is NLxNL<BR>
> N is small (5 or 6) and L is big (1e6)<BR>
><BR>
> One way is to build a loop and to compute L times the inversion of the<BR>
> NxN matrixes<BR>
> However, as L is big (for example 1e6) this is too much time consuming<BR>
><BR>
> Does anyone has an idea to compute this using scilab ?<BR>
><BR>
> Thank you for your help<BR>
><BR>
> Guylaine Collewet<BR>
> guylaine.collewet@cemagref.fr<BR>
> +33223482167<BR>
><BR>
><BR>
Stupid question first: do you have tested to store your big matrix in a<BR>
sparse matrix ?<BR>
stacksize('max')<BR>
B = spzeros(1e6,1e6);<BR>
// Store the matrix<BR>
Binv = inv(B);<BR>
<BR>
YC<BR>
<BR>
<BR>
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