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Eigenvectors are not orthogonal for some skew-symmetric matrices, why?

Asked by Rahul Singh on 1 May 2015
Latest activity Answered by Christine Tobler on 20 Sep 2018
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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3 Answers

Answer by Roger Stafford on 1 May 2015
Edited by Walter Roberson
on 20 Sep 2018
 Accepted Answer

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.


@Rahul: Once the repetition of eigenvalues is recognized, the adjustment can easily be done using matlab's 'orth' function. In the case of your particular A, the first and fourth eigenvalues in diag(E) are equal (on my computer), and similarly for the second and fifth, and for the third and sixth, so the adjustment (in my case) would simply be:
U(:,[1,4]) = orth(U(:,[1,4]));
U(:,[2,5]) = orth(U(:,[2,5]));
U(:,[3,6]) = orth(U(:,[3,6]));
Note that the resulting U, besides being orthonormal, will still constitute a valid set of eigenvectors, even though there is an arbitrary aspect to the result of 'orth'.
The hard part is recognizing such eigenvalue equalities and in fact recognizing matrices that are diagonalizable. There really are matrices that cannot be diagonalized and which are therefore designated as "defective". For these reasons it ought to be Mathworks that carries out such a revision.
However, you can experiment on your own using 'orth' to see how it works. Remember, both the eigenvalues and the eigenvectors will be complex-valued for your skew-symmetric matrices, and in testing the adjusted U'*U you will get tiny imaginary components due to rounding errors.
By the way, in requesting a change, you should probably not refer to the current version as being in "error", since Mathworks up to this point hasn't promised orthonormal eigenvectors for matrices that are other than real and symmetric.
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

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Answer by Rahul Singh on 2 May 2015

Is there any other way (other than Matlab) of computing orthogonal eigenvectors for this particular skew-symmetric matrix ? I tried NumPy package also, which gave me same results as Matlab.


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The adjusted U'*U and U*U' values using 'orth' are not significantly different on my computer. They differ only from the 6 x 6 identity matrix and from each other out at the 15-th or 16-th decimal place which is what can be expected from round-off errors. Correspondingly tiny imaginary parts are also present, again due to round-off errors. Are you sure you have like eigenvalues matched? Yours may have been given in a different order from mine.
Are the eigenvector pairs still orthogonal?
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.

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Answer by Christine Tobler on 20 Sep 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 =
>> norm(A*U - U*D)
ans =


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