eig function missing eigenvector

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David Mandel
David Mandel 2016-1-27
回答： Roger Stafford 2016-1-28
Consider the matrix
A = [0.5 0.5 0.0 0.0;
0.5 0.5 0.0 0.0;
0.0 0.0 0.5 0.5;
0.0 0.0 0.5 0.5];
This matrix clearly has an eigenvector of (1,1,1,1) with eigenvalue 1.
However, using the command
[V,D] = eig(A)
gives
V =
-0.7071 0 0 0.7071
0.7071 0 0 0.7071
0 -0.7071 0.7071 0
0 0.7071 0.7071 0
D =
0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 1
In particular, neither of the eigenvectors associated with the two eigenvalues of 1 are the (1,1,1,1) vector. Instead, it seems the eig function orthogonalized the set of eigenvectors. How can I recover the (1,1,1,1) eigenvector?

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回答（2 个）

Roger Stafford
Roger Stafford 2016-1-28
When you have more than one eigenvector with the same eigenvalue, any linear combination of them will also be an eigenvector with that same eigenvalue, of which there would be infinitely many, even if normalized. In your case the sum of those two rightmost eigenvectors times 1/sqrt(2) would give you the vector [1;1;1;1] which would also be an eigenvector. It cannot give you the infinitude of all possible eigenvectors with that eigenvalue. The function 'eig' simply chose a different pair of orthogonal eigenvectors than the particular one(s) you had in mind.
Walter Roberson
Walter Roberson 2016-1-27
eigenvectors are not unique, but they are certain to be orthogonal to each other for real symmetric matrices.

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