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 John Armstrong | Algebra, Linear Algebra

2 Comments »

[…] 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
[…] 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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