LU factorization for rectangular matrix seems incomplete
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The following example shows 3x4 maxtrix lu factorization in MATLAB seems incomplete.
MATLAB's answer for u is [1 3 3 2; 0 0 3 3; 0 0 6 6].
The answer in the text book (Gilbert Strang) is [1 3 3 2; 0 0 3 3; 0 0 0 0].
Is thera a way to get the same answer as the one from the text book through MATLAB?
----------------------------------------------------------------------------------------------------------
a =
[ 1, 3, 3, 2]
[ 2, 6, 9, 7]
[ -1, -3, 3, 4]
>> [l,u,p]=lu(a)
l =
[ 1, 0, 0]
[ 2, 1, 0]
[ -1, 0, 1]
u =
[ 1, 3, 3, 2]
[ 0, 0, 3, 3]
[ 0, 0, 6, 6]
p =
1 0 0
0 1 0
0 0 1
6 comentarios
Bruno Luong
el 18 de Nov. de 2018
Editada: Bruno Luong
el 18 de Nov. de 2018
Are you runing MATLAB or something else? Your formatting seems not to be MATLAB and your result does not match mine (R2018a).
a
a =
1 3 3 2
2 6 9 7
-1 -3 3 4
>> [l,u,p]=lu(a)
l =
1.000000000000000 0 0
0.500000000000000 1.000000000000000 0
-0.500000000000000 0 1.000000000000000
u =
2.000000000000000 6.000000000000000 9.000000000000000 7.000000000000000
0 0 -1.500000000000000 -1.500000000000000
0 0 7.500000000000000 7.500000000000000
p =
0 1 0
1 0 0
0 0 1
Soo Chang Choe
el 18 de Nov. de 2018
Bruno Luong
el 18 de Nov. de 2018
Why there is "problem" or "incomplete"? It's upper triangular as advertised.
For your information LU factorization is not unique.
Soo Chang Choe
el 19 de Nov. de 2018
Editada: Soo Chang Choe
el 19 de Nov. de 2018
Bruno Luong
el 19 de Nov. de 2018
Editada: Bruno Luong
el 19 de Nov. de 2018
"I don't understand why it stopped pivoting at the second row."
It does NOT stop, at the second step it pivots at the diagonal elements which is 0 for both row 2 and 3 (draw) so it picks row 2. Elements above/on the right the diagonal of u are not supposed to play any role.
In your book it selects row 3, but it is just a choise, without any additional usefulness for what it design for: solving linear system.
Both are equally valid.
Soo Chang Choe
el 19 de Nov. de 2018
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