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Eigenvectors - Eigenvalues

Nov 1, 2014 at 8:55 PM
Hallo all,

I am trying to use meta.numerics library for spectral graph analysis. More specific I am interested in inexact graph matching using the Eigenvectors. I have been testing the results of meta.numerics against the results of wolfram alpha (for small matrices).

The resulting Eigenvetors for unique Eigenvalues are corresponding but the resulting Eigenvectors for not unique Eigenvalues are totaly different. Unfrotunatly I am not sure if I comletly understood the math so maybe it is just me not understanding the math behind it.
In any case maybe somebody can find the time and have a look at it.

Laplacian Matrix:

4,0,0,-1,-1,-1,-1
0,4,0,-1,-1,-1,-1
0,0,4,-1,-1,-1,-1
-1,-1,-1,3,0,0,0
-1,-1,-1,0,3,0,0
-1,-1,-1,0,0,3,0
-1,-1,-1,0,0,0,3

Resulting Eigenvalues 7, 4, 4, 3, 3, 3, 0 (both wolfram alpha, meta.numerics)

Resulting Eigenvectors for Eigenvalues = 4 (normalized)

v2.m = (-0.566984,0.792311,-0.225327,0,0,0,0) (result meta.numerics)
v2.a = (-0.707107,0,0.707107,0,0,0,0) (result wolfram alpha)
v3.m = (-0.587534,-0.197256,0.784789,0,0,0,0) (result meta.numerics)
v3.a = (-0.707107,0.707107,0,0,0,0,0) (result wolfram alpha)


thx for any help

Richard
Coordinator
Feb 9, 2015 at 1:46 AM
Suppose two eigenvectors u and v have the same eigenvalue e. Then Au = eu and Av = ev. And by linearity, A(u + v) = e(u + v), and A(u - 0.5v) = e(u - 0.5)v, and in fact A(au + bv) = e(au + bv), i.e. any linear combination of those two eigenvectors is also an eigenvector with eigenvalue e. So there is actually a two-dimensional eigen-subspace for that eigenvalue, and any two vectors that span than subspace are as good an answer as any two others for "the" eigenvectors.

I just checked by hand up to three decimal places and {v2.m, v3.m} span the same subspace as {v2.a, v3.a}.