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From Ted Dunning <>
Subject Re: eigenvectors and eigenvalues of a matrix
Date Tue, 07 Jan 2014 06:57:13 GMT
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

    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

> 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

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