The Unapologetic Mathematician

Mathematics for the interested outsider

Repeated Eigenvalues

So we’ve got a linear transformation T on a vector space V of finite dimension d. We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are d 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 T.

But what if a root is repeated? Say the characteristic polynomial has a factor of (\lambda-1)^2. We might think to pick two linearly independent eigenvectors with eigenvalue {1}, 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 {1}. We can easily calculate its action on a vector


So the eigenvector equation says that x=x+y and y=y. The second is trivial, and the first tells us that y=0. 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.

February 11, 2009 - Posted by | Algebra, Linear Algebra


  1. […] 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 […]

    Pingback by Generalized Eigenvectors « The Unapologetic Mathematician | February 16, 2009 | Reply

  2. […] 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 […]

    Pingback by The Multiplicity of an Eigenvalue « The Unapologetic Mathematician | February 19, 2009 | Reply

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