Repeated Eigenvalues
So we’ve got a linear transformation 
on a vector space 
of finite dimension 
. We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be if we’re working over an algebraically closed field) then we can pick a basis of eigenvectors and get a diagonal matrix for .
But what if a root is repeated? Say the characteristic polynomial has a factor of 
. We might think to pick two linearly independent eigenvectors with eigenvalue 
, but unfortunately that doesn’t always work. Consider the transformation given by the matrix
This matrix is upper-triangular, and so we can just read off its eigenvalues from the diagonal: two copies of the eigenvalue . We can easily calculate its action on a vector
So the eigenvector equation says that 
and 
. The second is trivial, and the first tells us that 
. But we can only pick one linearly independent vector of this form. So we can’t find a basis of eigenvectors, and we can’t diagonalize this matrix. We’re going to need another approach.
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February 11, 2009 -
Posted by John Armstrong |
Algebra, Linear Algebra
[…] we’ve established that distinct eigenvalues allow us to diagonalize a matrix, but repeated eigenvalues cause us problems. We need to generalize the concept of eigenvectors […]
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[…] eigenvectors. That is, as the dimension of the kernel of . Unfortunately, we saw that when we have repeated eigenvalues, sometimes this doesn’t quite capture the right notion. In that example, the -eigenspace has […]
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