# LU decomposition

1. Oct 9, 2008

### dirk_mec1

Last edited by a moderator: May 3, 2017
2. Oct 9, 2008

### cellotim

Suppose we have two different LU decompositions, A = LU and A=L'U'. Because A is non-singular L, U, L' and U' are all non-singular and invertible. This implies that $$U = L^{-1}L'U'$$. Now you should be able to show that $$I = L^{-1}L'$$ in order to preserve upper-triangularity.

3. Oct 9, 2008

### dirk_mec1

Aha, very smart!

Well...from my instructors note the L is defined in a product of M-matrices(see picture) but if that's the case then L and L' have different M's (for we presume L and L' to be different).

If I invert a upper triangular matrix it remains upper triangular so I don't see what you mean by "preserving upper-triangularity".

4. Oct 9, 2008

### cellotim

The M's are not necessary as far as I can see only that L and L' are lower-triangular and U and U' are upper-triangular, and I mean preserve upper-triangularity of U' to U.

5. Oct 9, 2008

### dirk_mec1

So to preserve upper triangularity of U w.r.t. U' L-1 L' should be I?

6. Oct 9, 2008

### cellotim

It needs to be diagonal. It remains to prove that that diagonal matrix must be the identity.

Last edited: Oct 9, 2008
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