The order of the singular values and vectors should tell you.
For others who might be curious, the singular value decomposition breaks a
matrix A into three factors
A = U S V'
Both U and V are orthonormal so that U' U = I and V' V = I. S is diagonal.
An eigenvalue decomposition decomposes a square matrix into two parts, one
repeated
A = U S U*
If A is symmetric, then these are real matrices and U* = U'.
If A is symmetric, we can also decompose it using a Cholesky transformation
A = R' R
Where R is upper (right) triangular. We can then decompose R with SVD.
This gives us:
A = R' R
R = U S V'
R' R = (U S V')' (U S V') = V S U' U S V' = V S^2 V'
a nice convenience is that S^2 is also diagonal and contains the elements
of S, just squared.
So the answer for Tharindu is that the elements of V are not changed or
reordered and neither are the elements of S.
On Mon, Jan 6, 2014 at 10:22 PM, Tharindu Rusira
<tharindurusira@gmail.com>wrote:
> Hi,
> I am currently working with SingularValueDecomposition class and I like to
> clarify the following.
>
> My goal is to find eigenvalues and corresponding eigenvectors of a matrix.
> I know how to calculate eigenvalues and eigenvectors using svd but is there
> a way to keep track of which eigenvector corresponds to which eigenvalue?
>
> Thanks,
> 
> M.P. Tharindu Rusira Kumara
>
> Department of Computer Science and Engineering,
> University of Moratuwa,
> Sri Lanka.
> +94757033733
> www.tharindurusira.blogspot.com
>
