Eigenvectors are not orthogonal for some skew-symmetric matrices, why?
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Rahul Singh
il 1 Mag 2015
Commentato: Paul
il 16 Set 2025 alle 2:00
0 -0.5000 0 0 0 0.5000
0.5000 0 -0.5000 0 0 0
0 0.5000 0 -0.5000 0 0
A = 0 0 0.5000 0 -0.5000 0
0 0 0 0.5000 0 -0.5000
-0.5000 0 0 0 0.5000 0
The above matrix is skew-symmetric. When I use [U E] = eig(A), to find the eigenvectors of the matrix. These eigenvectors must be orthogonal, i.e., U*U' matix must be Identity matrix. However, I am getting U*U' as
0.9855 -0.0000 0.0410 -0.0000 -0.0265 0.0000
-0.0000 0.9590 0.0000 0.0265 -0.0000 0.0145
0.0410 0.0000 0.9735 -0.0000 -0.0145 0.0000
-0.0000 0.0265 -0.0000 1.0145 0.0000 -0.0410
-0.0265 -0.0000 -0.0145 0.0000 1.0410 -0.0000
0.0000 0.0145 0.0000 -0.0410 -0.0000 1.0265
Here we can observe a substantial error. This happens for some other skew-symmetric matrices also. Why this large error is being observed and how do I get correct eigen-decomposition for all skew-symmetric matrices?
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Risposta accettata
Roger Stafford
il 1 Mag 2015
Modificato: Walter Roberson
il 20 Set 2018
Your matrix A is "defective" , meaning that its eigenvalues are not all distinct. In fact, it has only three distinct eigenvalues. Consequently the space of eigenvectors does not fully span six-dimensional vector space. See the Wikipedia article:
What you are seeing is not an error on Matlab's part. It is a mathematical property of such matrices. You cannot achieve what you call "correct eigen-decomposition" for such matrices.
8 Commenti
Lorenzo
il 20 Set 2018
Modificato: Lorenzo
il 20 Set 2018
That matrix is not defective (1i times the matrix is hermitian and so it has a complete set of eigenvectors), it has however degenerate eigenvalues and this is the reason why U fails to be unitary. You should use schur in this case, which always return a unitary matrix
Più risposte (2)
Rahul Singh
il 2 Mag 2015
4 Commenti
Roger Stafford
il 3 Mag 2015
Yes, all the eigenvectors come out orthogonal after that adjustment I described. The fact that U'*U gives the identity matrix implies that. You should be able to check that for yourself.
Christine Tobler
il 20 Set 2018
Since, as Lorenzo points out in a comment above, 1i*A is hermitian, you could apply eig to that matrix:
>> [U, D] = eig(1i*A);
>> D = D/1i;
>> norm(U'*U - eye(6))
ans =
1.4373e-15
>> norm(A*U - U*D)
ans =
7.8098e-16
1 Commento
Paul
il 16 Set 2025 alle 2:00
"As of R2021a, eig will now detect if a matrix is skew-symmetric or skew-hermitian, and will return purely imaginary eigenvalues and orthogonal eigenvectors in that case. See the Release Notes. This does require the input to be exactly skew-hermitian, not just up to round-off (you can use ishermitian(A, 'skew') to check this)." Source
A = [ 0 -0.5000 0 0 0 0.5000
0.5000 0 -0.5000 0 0 0
0 0.5000 0 -0.5000 0 0
0 0 0.5000 0 -0.5000 0
0 0 0 0.5000 0 -0.5000
-0.5000 0 0 0 0.5000 0];
[U,D] = eig(A,'vector');
any(real(D))
norm(U'*U - eye(6))
norm(U*U' - eye(6))
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