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Ajuda do Scilab >> Interfaces com UMFPACK (sparse) > condestsp

condestsp

estimate the condition number of a sparse matrix

Calling Sequence

[K1] = condestsp(A, LUp, t)
[K1] = condestsp(A, LUp)
[K1] = condestsp(A, t)
[K1] = condestsp(A)

Arguments

A

a real or complex square sparse matrix

LUp

(optional) a pointer to (umf) LU factors of A obtained by a call to umf_lufact ; if you have already computed the LU (= PAQ) factors it is recommended to give this optional parameter (as the factorization may be time consuming)

t

(optional) a positive integer (default value 2) by increasing this one you may hope to get a better (even exact) estimate

K1

estimated 1-norm condition number of A

Description

Give an estimate of the 1-norm condition number of the sparse matrix A by Algorithm 2.4 appearing in :

"A block algorithm for matrix 1-norm estimation
 with an application to 1-norm pseudospectra"
 Nicholas J. Higham and Francoise Tisseur
 Siam J. Matrix Anal. Appl., vol 21, No 4, pp 1185-1201

Noting the exact condition number K1e = ||A||_1 ||A^(-1)||_1, we have always K1 <= K1e and this estimate gives in most case something superior to 1/2 K1e

Examples

A = sparse( [ 2  3  0  0  0;
              3  0  4  0  6; 
              0 -1 -3  2  0; 
              0  0  1  0  0; 
              0  4  2  0  1] );
K1 = condestsp(A)
// verif by direct computation
K1e = norm(A,1)*norm(inv(full(A)),1)

// another example
[A] = ReadHBSparse(SCI+"/modules/umfpack/examples/arc130.rua");
K1 = condestsp(A)
// this example is not so big so that we can do the verif
K1e = norm(A,1)*norm(inv(full(A)),1)

// if you have already the lu factors condestsp(A,Lup) is faster
// because lu factors are then not computed inside condestsp
Lup = umf_lufact(A);   
K1 = condestsp(A,Lup)
umf_ludel(Lup)         // clear memory

See Also

  • umf_lufact — lu factorization of a sparse matrix
  • rcond — número de condicionamento inverso
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Last updated:
Fri Apr 11 14:18:13 CEST 2014