Re: Problem with Singular Value Decomposition Algorithm

[Brian Gough <bjg at network-theory dot co dot uk>]

Columns 2 and 3 have the opposite sign, but this is the arbitrary
minus-sign factor referred to earlier.  The results satisfy m =
u*diag(s)*v' and u'*u = I, v'*v = I numerically --- so the
decomposition looks ok.  Let me know if there's something I missing
here.

regards
Brian Gough

Jim Love writes:
 > This code fixes the order problem of the S vector and the other matrix, but their is still a sign problem. Using this matrix for A:
 > 
 > 1.000000 1.000000 0.975000
 > 1.000000 -1.000000 0.975000
 > -1.000000 -1.000000 -0.925000
 > -1.000000 1.000000 -1.025000
 > 
 > The modified code produces:
 > 
 > s:
 > 2.793961 2.000000 0.035791
 > 
 > This is correct!
 > 
 > For V:
 > 
 > -0.715538 -0.025633 0.698103
 > 0.018347 -0.999671 -0.017900
 > -0.698332 -0.000000 -0.715774
 > 
 > This is NOT correct! 
 > 
 > The correct answer for V is:
 > 
 > -0.7155    0.0256   -0.6981
 >     0.0183    0.9997    0.0179
 >    -0.6983   -0.0000    0.7158
 > 
 > U is also wrong:  the program outputs:
 > 
 > -0.493230 -0.512652 -0.493875
 > -0.506363 0.487019 0.506368
 > 0.480733 0.512652 -0.506047
 > 0.518861 -0.487019 0.493554
 > 
 > The correct U is:
 > 
 > -0.4932    0.5127    0.4939    
 >    -0.5064   -0.4870   -0.5064   
 >     0.4807   -0.5127    0.5060   
 >     0.5189    0.4870   -0.4936   
 > 
 > Note last column missing for both solutions  for U.
 >