Which SVD algorithm implementation is used in MATLAB?

Question

Danijel Domazet 2022-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/1642020-which-svd-algorithm-implementation-is-used-in-matlab

评论： Walter Roberson 2022-2-22

Hi,

I am comparing singular value decomposition function [U,S,V] = svd(A) to some C implementations of the algorithm. However, I am getting somewhat different results: for example, columns of the output matrix U are mixed-up, or some output values have different sign, etc.

Does anyone know which SVD algorithm implementation is used in MATLAB?

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Danijel Domazet 2022-2-22

编辑：Danijel Domazet 2022-2-22

Walter, I have tried LAPACK, but no success. Please check here:

https://www.mathworks.com/matlabcentral/answers/1655690-matlab-and-lapack-implementation-of-the-svd-algorithm-not-the-same-output.

Walter Roberson 2022-2-22

I am not sure how it is relevant that MKL is copyrighted? LAPACK is copyrighted too, with a modified BSD license.

MKL is probably closed source, but you did not indicate that you needed source access; for everything you have mentioned so far you only need to be able to link against it.

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Answer 1

Steven Lord 2022-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/1642020-which-svd-algorithm-implementation-is-used-in-matlab#answer_888260

在 MATLAB Online 中打开

Keep in mind that the SVD of a matrix is not unique. Quoting from Wikipedia: "Non-degenerate singular values always have unique left- and right-singular vectors, up to multiplication by a unit-phase factor

(for the real case up to a sign). Consequently, if all singular values of a square matrix M are non-degenerate and non-zero, then its singular value decomposition is unique, up to multiplication of a column of U by a unit-phase factor and simultaneous multiplication of the corresponding column of V by the same unit-phase factor."

A = magic(5);
[U, S, V] = svd(A);
format longg
check1 = U*S*V' - A
check1 = 5×5
	1.0e+00 *

       3.5527136788005e-15      -1.4210854715202e-14     -1.93178806284777e-14      2.48689957516035e-14      1.59872115546023e-14
                         0     -1.77635683940025e-14      -1.4210854715202e-14      2.48689957516035e-14      1.77635683940025e-14
      -3.5527136788005e-15      6.21724893790088e-15      1.77635683940025e-14        7.105427357601e-15        7.105427357601e-15
      -3.5527136788005e-15      3.19744231092045e-14      1.77635683940025e-14     -1.77635683940025e-14     -8.43769498715119e-15
      5.32907051820075e-15       3.5527136788005e-14       1.4210854715202e-14     -2.17603712826531e-14     -1.95399252334028e-14
U2 = -U;
V2 = -V;
check2 = U2*S*V2' - A
check2 = 5×5
	1.0e+00 *

       3.5527136788005e-15      -1.4210854715202e-14     -1.93178806284777e-14      2.48689957516035e-14      1.59872115546023e-14
                         0     -1.77635683940025e-14      -1.4210854715202e-14      2.48689957516035e-14      1.77635683940025e-14
      -3.5527136788005e-15      6.21724893790088e-15      1.77635683940025e-14        7.105427357601e-15        7.105427357601e-15
      -3.5527136788005e-15      3.19744231092045e-14      1.77635683940025e-14     -1.77635683940025e-14     -8.43769498715119e-15
      5.32907051820075e-15       3.5527136788005e-14       1.4210854715202e-14     -2.17603712826531e-14     -1.95399252334028e-14

The elements of check1 and check2 both look sufficiently small as to be close enough to 0 to say both (U, S, V) and (U2, S, V2) satisfy the SVD contract. It's possible the columns of the output U aren't "mixed up" they're just a different (correct) answer (when combined with the corresponding V) than you expected.