luļ
Syntax
lu(obj, [permute=false])
Arguments
obj is a matrix with no NULL values.
permute is a Boolean value. The default value is false.
Details
Compute pivoted LU decomposition of a matrix.
If permute is false, return 3 matrices (L, U and P) with obj = PāLU. P is a permutation matrix, L is a lower triangular matrix with unit diagonal elements, and U is an upper triangular matrix.
If permute is true, return 2 matrices (L and U) with obj = L*U.
Examples
$ A = matrix([[2, 5, 8, 7], [5, 2, 2, 8], [7, 5, 6, 6], [5, 4, 4, 8]]);
$ P, L, U = lu(A);
$ P;
#0 |
#1 |
#2 |
#3 |
|---|---|---|---|
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
$ L;
#0 |
#1 |
#2 |
#3 |
|---|---|---|---|
1 |
0 |
0 |
0 |
0.875 |
1 |
0 |
0 |
0.25 |
0.72 |
1 |
0 |
0.625 |
0.12 |
0.233871 |
1 |
$ U;
#0 |
#1 |
#2 |
#3 |
|---|---|---|---|
8 |
2 |
6 |
4 |
0 |
6.25 |
0.75 |
4.5 |
0 |
0 |
4.96 |
0.76 |
0 |
0 |
0 |
0.782258 |
$ L, U = lu(A, true);
$ L;
#0 |
#1 |
#2 |
#3 |
|---|---|---|---|
0.25 |
0.72 |
1 |
0 |
0.625 |
0.12 |
0.233871 |
1 |
1 |
0 |
0 |
0 |
0.875 |
1 |
0 |
0 |
$ U;
#0 |
#1 |
#2 |
#3 |
|---|---|---|---|
8 |
2 |
6 |
4 |
0 |
6.25 |
0.75 |
4.5 |
0 |
0 |
4.96 |
0.76 |
0 |
0 |
0 |
0.782258 |