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

\displaystyle\begin{pmatrix}1&1\\{0}&1\end{pmatrix}

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

\displaystyle\begin{pmatrix}1&1\\{0}&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\y\end{pmatrix}

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 | 2 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 391 other followers