eigenvectors and spectral decomposition

osigrtoeltParticipant 2010-12-17 at 2:50 am

How do we do spectral decomposition (eigendecomposition) of a Hermitian matrix? I wasn’t able to get the orthonormal basis of eigenvectors if the characteristic equation has multiple roots. Thanks.

In R, we can do something like this:

> x < - diag(rep(1, 4)) > eigen(x)

$values [1] 1 1 1 1

$vectors [,1] [,2] [,3] [,4] [1,] 0 0 0 1 [2,] 0 0 1 0 [3,] 0 1 0 0 [4,] 1 0 0 0

Haksun LiModerator 2010-12-17 at 5:47 am

<br />
       Matrix A = new DiagonalMatrix(R.rep(1.0, 4));<br />
<br />
       Eigen instance = new Eigen(A, Eigen.Method.QR, 0);<br />
       Property prop = instance.getProperty(0);<br />
       assertEquals(NumberUtil.parse("1.0"), instance.eigenvalue(0));<br />
       assertEquals(4, prop.geometricMultiplicity());<br />
       VectorList basis = instance.getProperty(0).eigenbasis();<br />

[tt:q1tj2dox]basis[/tt:q1tj2dox] has all the 4 vectors you are looking for.

osigrtoeltParticipant 2010-12-17 at 8:06 am

Thanks for the answer, Haksun. However, I still don’t see how this should be done for a more general matrix with eigenvalues of multiplicity > 1.

Can we please implement the eigenvalue decomposition as a class like Cholesky or SVD? For a symmetric matrix A, we should have A = QDQ^{-1} = QDQ^T, where D is a diagonal matrix that contains all eigenvalues of A (counting multiplicity) and Q is an orthonormal matrix.

http://en.wikipedia.org/wiki/Eigenvalue_decomposition
Haksun LiModerator 2010-12-17 at 9:40 am

It is already there.

This example has only one eigenvalue with multiplicity of 4.

The dimension of the space spanned by its basis is therefore 4. So, there are 4 orthogonal basis vector. Any linear combination of these 4 vector is a eigenvector.

To get the D, you just form a diagonal matrix using the eigenvalues, e.g., by new DiagonalMatrix(e1, e2,…). To get the Q, you just form a matrix using the (basis) eigen vector, e.g., using R.cbind(v1, v2, v3, v4, …)

We will output D and Q more explicitly in the next release.

Haksun LiModerator 2010-12-17 at 1:46 pm

We now have eigen decomposition, e.g.,

<br />
        DenseMatrix A = new DenseMatrix(<br />
                new SymmetricMatrix(new double[][]{<br />
                    {1,},<br />
                    {2, 3},<br />
                    {4, 5, 6},<br />
                    {7, 8, 9, 10}<br />
                }));<br />
        Eigen eigen = new Eigen(A, Eigen.Method.QR, 1e-4);//TOOD: it is important that we choose a smaller precision<br />
<br />
        Matrix D = eigen.D();<br />
        Matrix Q = eigen.Q();<br />
        Matrix Qt = eigen.Qt();<br />
<br />
        Matrix B = Q.multiply(D).multiply(Qt);<br />
        AreMatrices.equal(A, B, 1e-14);<br />

ilyashParticipant 2013-02-11 at 4:00 pm

I decided to give a hand and sent a post into social bookmarks. I hope the popularity will rise in.
Haksun LiModerator 2013-02-12 at 3:00 am

Thank you, CamKrist. It is appreciated.