Eigenvectors with Identical Eigenvalues Lose orthogonality?
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From: NIcholas on 6 Aug 2010 12:22 Shouldn't the command "[V,D] = eig(A);" give an orthogonal basis V? If A has unique eigenvalues I find this to be true, but if A has identical eigenvalues (see code below) the eigenvectors with identical eigenvalues seem to lose orthogonality. Is this a numerical issue with the eig command? and is there a way around it? CODE: rand('state',0); [Q,R]=qr(rand(4)); % Q is a random orthogonal basis A=Q * diag([1 2 3 3]) * Q'; % eigenvalue 3 has multiplicity 2 [V,D]=eig(A); disp(V'*V) 1.0000 0.0000 -0.0000 -0.0000 0.0000 1.0000 0.0000 -0.0000 -0.0000 0.0000 1.0000 -0.3161 -0.0000 -0.0000 -0.3161 1.0000 If instead I give A unique eigenvalues I do get an orthogonal basis: A=Q * diag([1 2 3 4]) * Q'; [V,D]=eig(A); disp(V'*V) 1.0000 0.0000 0.0000 0.0000 0.0000 1.0000 -0.0000 0.0000 0.0000 -0.0000 1.0000 0.0000 0.0000 0.0000 0.0000 1.0000
From: Matt J on 6 Aug 2010 12:54 "NIcholas " <dsfadfa(a)aol.com> wrote in message <i3hcrc$g9o$1(a)fred.mathworks.com>... > Shouldn't the command "[V,D] = eig(A);" give an orthogonal basis V? ============== If A is not Hermitian, it might not even have a fulll set of linear independent eigenvectors, making it impossible for eig() to produce an orthogonal basis. In your example, A is Hermitian, but since eig doesn't know this in advance, it has no reason to try to orthogonalize the result.
From: Matt J on 6 Aug 2010 12:59 "NIcholas " <dsfadfa(a)aol.com> wrote in message <i3hcrc$g9o$1(a)fred.mathworks.com>... > If instead I give A unique eigenvalues I do get an orthogonal basis: ========== Yes, the eigenvectors of a Hermitian matrix belonging to different eigenvalues are always orthogonal to each other. A repeated eigenvalue with multiplicity M has an M-dimensional space of eigenvectors. You can orthogonalize any basis for this M-dimensional space using Gram-Schmidt, but again eig() has no reason to add this processing, not knowing whether A is Hermitian.
From: Matt J on 6 Aug 2010 13:16 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i3hf0q$3vk$1(a)fred.mathworks.com>... > A repeated eigenvalue with multiplicity M has an M-dimensional space of eigenvectors. You can orthogonalize any basis for this M-dimensional space using Gram-Schmidt, but again eig() has no reason to add this processing, not knowing whether A is Hermitian. ================ eig() also has no way of detecting whether eigenvalues are repeated. The test EigenValue1==EigenValue2 is too naive to account for floating point errors in the computation of the eigenvalues.
From: Bruno Luong on 6 Aug 2010 13:37
"NIcholas " <dsfadfa(a)aol.com> wrote in message <i3hcrc$g9o$1(a)fred.mathworks.com>... > Shouldn't the command "[V,D] = eig(A);" give an orthogonal basis V? > If A has unique eigenvalues I find this to be true, but if A has identical eigenvalues (see code below) the eigenvectors with identical eigenvalues seem to lose orthogonality. Is this a numerical issue with the eig command? and is there a way around it? This is not an "issue": there is no warranty that EIG must return the orthogonal basis for multiple-order eigenspace even for Hermitian matrix. Any basis vectors within the multilple-order eigenspace are still *valid* eigen-vectors. If you want to reorthonalize, use ORTH after EIG. Or you can call SVD [V D V] = svd(A) Bruno |