Hi Ted, Thank a lot for your comprehensive reply.
The following is a simple piece of code I wrote. According to what you have
said, this code would make sense, am I right? Please correct me if I'm
wrong.
/*respective parameters of the following method are taken from
getSingularvalues() and getV() methods in SingularValueDecomposition class.
*/
public HashMap<Double,DenseVector> mapEigenvaluesToEigenVectors(double[]
singularValues, Matrix eigenVectors) {
HashMap<Double,DenseVector> eigenMap = new
HashMap<>(singularValues.length);
for (int i = 0; i < singularValues.length; i++) {
// eigenvalue = singular value ^2
// eigenvectors = right singular vectors
eigenMap.put(singularValues[i] * singularValues[i], (DenseVector)
eigenVectors.viewRow(i));
}
return eigenMap;
}
In case my problem is unclear, here's some context,
I have an input matrix X and I want to find eigenvalues/eigenvectors of the
covariance matrix XTX. So my workaround is to find singular values and
right singular vector of X in order to use the following equivalency.
" Comparison with the eigenvector factorisation of *X*T*X* establishes that
the right singular vectors *W* of *X* are equivalent to the eigenvectors of
*X*T*X*, while the singular values *σ*(*k*) of *X* are equal to the square
roots of the eigenvalues *λ*(*k*) of *X*T*X*. " ~ Wikipedia [1]
[1]
http://en.wikipedia.org/wiki/Principal_component_analysis#Singular_value_decomposition
Regards,
On Tue, Jan 7, 2014 at 12:27 PM, Ted Dunning <[email protected]> wrote:
> 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
> re-ordered and neither are the elements of S.
>
>
> On Mon, Jan 6, 2014 at 10:22 PM, Tharindu Rusira
> <[email protected]>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.tharindu-rusira.blogspot.com
> >
>
--
M.P. Tharindu Rusira Kumara
Department of Computer Science and Engineering,
University of Moratuwa,
Sri Lanka.
+94757033733
www.tharindu-rusira.blogspot.com
