1
$\begingroup$

By doing some numerical examples I have satisfied myslef that the left and right singular vectors are equal for such matrices.

http://people.revoledu.com/kardi/tutorial/LinearAlgebra/SVD.html

However I would like to see a proof of this mathematically.

Also it seems to contradict the definintion given here:

http://en.wikipedia.org/wiki/Singular_value_decomposition

which says that:

"Non-degenerate singular values always have unique left- and right-singular vectors, up to multiplication by a unit-phase factor eiφ (for the real case up to sign). Consequently, if all singular values of 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."

So how come for a correlation matrix the svd can have equal right and left singular vectors without becoming degenerate?

Baz

$\endgroup$
3
$\begingroup$

Because a correlation matrix is symmetric , the right- and leftmultiplication to SVD/diagonalization are just the transposes of each other. Note, that "unique" does not mean "each one is different" here, but rather: we'll "unambiguously find a definitive solution" for the right- as well for the leftmultiplication.

$\endgroup$
1
$\begingroup$

The finite-dimensional version of the "spectral theorem" says every real symmetric matrix can be diagonalized by an orthogonal matrix. The singular value decomposition says every real matrix can be diagonalized by two orthogonal matrices. When it's symmetric, you get only one of them. It appears on the left and its transpose on the right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.