Re: [femm] triangle

[dcm3c@xxxxxxx]

In a message dated 10/24/2001 10:02:35 PM Eastern Daylight Time, sxy28@xxxxxxxxxxx writes:


> To Dear All:
>
> Hop are you? I would like to ask one more thing about re-ordering things. 
> There is an equation Ax = B we want to solve. After re-ordering the sparse 
> matrix A, the matrix becomes banded matrix A' Then the equation we want to 
> solve is that A'x = B. Should not we need to do something to matrix B?

Well, if you've already built an a matrix A and a vector B, you could think of the reordering as the mapping: z = P*x  where P is a permutation matrix.  That is, it's just the identity matrix with the rows shifted around, corresponding to how you want to reorder the matrix. These kind of matrices have the nice property that their transpose is their inverse.  Now, subsititute for x in terms of z and multiply both sides by P (to get a symmetric result) to get:

P * A * P^T z = P * B

This isn't a suprising result--it just means you have to reorder the RHS too.

It should be noted, however, that you almost never have to do these sort of matrix multiplies explicitly in the context of a finite element program.  You generally do the renumbering before you build the matrices.

Dave.
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