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

30 views (last 30 days)
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?

Accepted Answer

Roger Stafford
Roger Stafford on 1 May 2015
Edited: Walter Roberson on 20 Sep 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 Comments
Lorenzo
Lorenzo on 20 Sep 2018
Edited: Lorenzo on 20 Sep 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

Sign in to comment.

More Answers (2)

Rahul Singh
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.
  4 Comments
Roger Stafford
Roger Stafford on 3 May 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.

Sign in to comment.


Christine Tobler
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 =
1.4373e-15
>> norm(A*U - U*D)
ans =
7.8098e-16

Categories

Find more on Linear Algebra in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!