# The Unapologetic Mathematician

## 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