SVD of symmetric complex matrix is giving an unexpected result.

Suppose I have a symmetric complex matrix A. My understanding is that the the right singular vectors of this matrix should be equal to the complex conjugate of the left singular vectors. However, when I perform the SVD this is not what I am finding. Any help?

clear; clc;
g = randn(5,1) + 1i*randn(5,1);
K = g*g.';     % rank-1 symmetric complex matrix
[U,S,V] = svd(K);    
u1 = U(:,1);     % left singular vector
v1 = V(:,1);     % right singular vector
diff = u1 - conj(v1)     % I expect this to be close to zero

Answer 1

[Alfonso Nieto-Castanon]

Well, yes, they are "equal" up to a constant pure-phase term, which is always arbitrary in SVD (for any left and right singular vectors u and v, u*exp(1i*phi) and v*exp(1i*phi) would solve the SVD problem just as well, i.e. U*S*V'==K with U,V unitary and S nonnegative diagonal).

In this case the following diffU will be approximately zero:

 ph = angle(v1(1))+angle(u1(1));
 diffU = u1 - conj(v1*exp(-1i*ph));

More generally, for a matrix K with rank>1:

 g = randn(5)+1i*randn(5); 
 K = g*g.'; 
 [U,S,V] = svd(K); 
 ph = angle(U(1,:))+angle(V(1,:));
 V = V*diag(exp(-1i*ph)); 
 diffU = U-conj(V);

will also be approximately zero.

Answer 2

[John D'Errico]

Your problem is that there is a difference between .' and ' as operators. I'm not sure why you chose to use .' there, but it was perhaps a poor choice. :)

[Landon Chad]

I chose to use the .' operator because I wanted to create a symmetric complex matrix. If I had used the ' operator then I would end up with a Hermitian matrix. My understanding is that symmetric complex matrices should be factorable using SVD where the left singular vectors should be equal to the complex conjugate of the right singular vectors.

[Walter Roberson]

I did cross-check and the K being created is symmetric even at the bit level, being exactly equal to (K+K.')/2

[Landon Chad]

@WalterRoberson: So is there any particular reason why the left singular vectors are not equal to the complex conjugate of the right singular vectors? I thought that the SVD of a symmetric complex matrix should have left singular vectors being equal to the complex conjugate of the right singular vectors?

[Walter Roberson]

Sorry, linear algebra is a weakness of mine. I took the course 3 times and still have problems with it.

[John D'Errico]

Sigh. Choosing to use the wrong operator will make the result invalid. You can't just change operators randomly and expect mathematics to work.

For example, as long as a is finite and non-zero, we expect that a/a == 1.

However, would you also expect a*a == 1? Hey, I just swapped in a different operator. Whats the problem?

Your "understanding" about this result is simply wrong, IF you build K as you did, using the .' operator.

As I point out, IF you use K = g*g', then not surprisingly to me, you get the result that you claimed to have expected.

I think what you are both missing is that a complex symmetric matrix has the property that A==A', where ' is the ctranspose operator, thus the complex CONJUGATE transpose. You simply cannot use .' here when you form K. (People get used to working with real numbers, and then get sloppy about things like transposes.)