Hello

I need to find L-1 from the Cholesky decompostion
A = L . Ltranspose
and have implemented choldc from nr2 in C with additional lines to find inverse as given in chapter 2.9.

I have calculated L(L-1) as a test and do not get the identity matrix, I was just wondering whether anyone else had encountered this problem... or could point out if I'm being stupid and missing something?

Thanks
jbd

... or could point out ... Is it a multiple-choice test?

But seriously, what did you use as a test case? How did you get all of the elements of the calculated matrix L and L-inverse from the return values of choldc? How close did the answer come to expected?

Using choldc (single precision floating point) from nr2 C, here is some output from a matrix used as a test case in xcholdc, where I have added the stuff that I think you are asking about:


Original matrix:
100.000000 15.000000 0.010000
15.000000 2.300000 0.010000
0.010000 0.010000 1.000000

After choldc, diagonal elements of L and L-Transpose are in [p]:
10.000000
0.223607
0.999277

Upper of [a] contains the upper of the original matrix.
Below main diagonal of [a] is below main diagonal of L:
100.000000 15.000000 0.010000
1.500000 2.300000 0.010000
0.001000 0.038013 1.000000

Extracting L and L-Transpose from [p] and lower [a]:
L:
10.000000 0.000000 0.000000
1.500000 0.223607 0.000000
0.001000 0.038013 0.999277
L-Transpose:
10.000000 1.500000 0.001000
0.000000 0.223607 0.038013
0.000000 0.000000 0.999277

Product of L and L-Transpose should be equal to original:
100.000000 15.000000 0.010000
15.000000 2.300000 0.010000
0.010000 0.010000 1.000000

Inverse of L from code at end of section 2.9:
0.100000 0.000000 0.000000
-0.670821 4.472138 0.000000
0.025418 -0.170123 1.000724


Product of L and Inverse of L:
1.000000 0.000000 0.000000
0.000000 1.000000 0.000000
0.000000 0.000000 1.000000


Regards,

Dave