LU=A
11-22-2015, 04:24 PM (This post was last modified: 11-22-2015 04:34 PM by gipa.)
Post: #1
 gipa Junior Member Posts: 1 Joined: Nov 2015
LU=A
Hi Everybody,

I have a problem with the LU decomposition. It uses a P permutation matrix (LU factorization with Partial Pivoting) for the computing. It means that P*A=L*U. There are some cases when a proper permutation in rows (or columns) is necessary for the LU factorization. But what in those cases when it is unnecessary? There are some matrices, which can be solved with A=L*U without P permutation matrix.

For example:

A=
2 1 1
4 1 0
-2 2 1

L=
1 0 0
2 1 0
-1 -3 1

U=
2 1 1
0 -1 -2
0 0 -4

If I try to solve it with HP Prime I get the following solution:

L=
1 0 0
-0.5 1 0
0.5 0.2 1

U=
4 1 0
0 2.5 1
0 0 0.8

and it uses a
P=
0 1 0
0 0 1
1 0 0

Theorem: Any square matrix A admits an LUP factorization. If A is invertible, then it admits an LU factorization if and only if all its leading principal minors are nonsingular. If LU factorization exists it means that it is unique.

I think it means that the computer just does LUP factorization in the case of every matrices without checking invertibility. But it gives an LU factorization for another matrix instead of what I need.

Can anybody help me to solve this problem?
11-22-2015, 05:17 PM
Post: #2
 parisse Senior Member Posts: 1,101 Joined: Dec 2013
RE: LU=A
There is a non trivial P for numerical stability if data is approx. Try lu in CAS. with exact data instead
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