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
Best Answer
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.