Linear algebra of operators

[JohanL]

If you take the outer product of two orthogonal 4-dimensional vectors you get a 4*4 matrix. I wonder what you generally can say about this matrix eigenvectors? For example: If 2,3 or 4 of the eigenvalues are zero is it always true that the matrix have at least 2 linear dependent eigenvectors? Maybe this is true regardless of the values of the eigenvalues?

The physical problem is about the operator |a><b| where |a> and |b> are orthogonal to each other and the characteristic polynom is x^4. I have the values for this operator in a basis and it can not be diagonalized.
Is this maybe a general property when |a> and |b> are orthogonal? Or when a couple of the eigenvalues are zero?
If so what is the physics behind this?

Any ideas?

 

[lippyka]

Yes, it is generally true that if two, three, or four of the eigenvalues are zero, then the matrix has at least two linearly dependent eigenvectors. This is because any matrix with a dimension greater than the rank of the matrix must contain linearly dependent eigenvectors. In this case, the rank of the 4x4 matrix is less than 4, since at least two of the eigenvalues are zero. The physics behind this is that when two orthogonal vectors are multiplied together, they form a matrix whose eigenvalues will be zero, as the two vectors do not have any overlap. Therefore, the matrix will necessarily have at least two linearly dependent eigenvectors.