1. ## Basis of eigenvectors

Just a quick Q -- I know how to find the characteristic polynomial, solve for the eigenvalues, and find the eigenvectors from there, but I was wondering, say you have a double eigenvalue (like your polynomial reduces to (x-5)(x-5) -- then you would have a double eigenvalue of 5)...is it then possible to determine a basis of eigenvalues for a 2x2 matrix? Or since you only have one eigenvector, can you not determine a basis?

2. Originally Posted by scosgurl
Just a quick Q -- I know how to find the characteristic polynomial, solve for the eigenvalues, and find the eigenvectors from there, but I was wondering, say you have a double eigenvalue (like your polynomial reduces to (x-5)(x-5) -- then you would have a double eigenvalue of 5)...is it then possible to determine a basis of eigenvalues for a 2x2 matrix? Or since you only have one eigenvector, can you not determine a basis?
If there is a double eigenvalue, then it may or may not have two linearly independent eigenvectors (and similarly a triple eigenvalue can have one, two, or three, linearly independent eigenvectors, and so on). If there are "enough" lin. ind. eigenvectors for each eigenvalue, then the matrix will be diagonalisable. otherwise, it will not be.