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On page 18 of SuperLU's user guide, the following sample input matrix A is given:

19,0,21,21,0
12,21,0,0,0
0,12,16,0,0
0,0,0,5,21
12,12,0,0,18

U matrix is given as follow in the user guide:

19,0,21,21,0
0,21,-13.26,-13.26,0
0,0,23.58,7,58,0
0,0,0,5,21
0,0,0,0,34.2

But dgssv() produces the following matrix, which is printed via dPrint_CompCol_Matrix("U", U)

CompCol matrix U:
Stype 0, Dtype 1, Mtype 4
nrow 5, ncol 5, nnz 11
nzval: 21.000000  -13.263158  7.578947  21.000000  
rowind: 0  1  2  0  
colptr: 0  0  0  1  4  4

The U matrix given in the user guide is correct (you can verify it at this link). I wonder why SuperLU produces such a matrix. How can I generate correct U matrix from the SuperLU output?

P.S. For ordering columns of A, I have used my permutation vector, which is 1,2,3,4,5. In other words, there is no reordering for columns. Rows of A are not reordered by SuperLU.

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I have found out that the missing nonzeros are stored in L matrix. Storage type of L is supernodel to speedup the factorization process via utilizing dense blocks. So some nonzeros of U matrix is stored in L matrix.

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  • $\begingroup$ Welcome to SciComp. No problem in answering your own questions, but for the sake of clarity for future visitors of the site, could you please write a more pedagogic answer? Thanks. $\endgroup$ – GertVdE Jul 8 '13 at 5:35

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