The Unapologetic Mathematician

Mathematics for the interested outsider

Generalized Eigenvectors

Sorry for the delay, but exam time is upon us, or at least my college algebra class.

Anyhow, 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 somewhat.

First of all, since an eigenspace generalizes a kernel, let’s consider a situation where we repeat the eigenvalue {0}:

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

This kills off the vector \begin{pmatrix}1\\{0}\end{pmatrix} right away. But the vector \begin{pmatrix}{0}\\1\end{pmatrix} gets sent to \begin{pmatrix}1\\{0}\end{pmatrix}, where it can be killed by a second application of the matrix. So while there may not be two independent eigenvectors with eigenvalue {0}, there can be another vector that is eventually killed off by repeated applications of the matrix.

More generally, consider a strictly upper-triangular matrix, all of whose diagonal entries are zero as well:

\displaystyle\begin{pmatrix}{0}&&*\\&\ddots&\\{0}&&{0}\end{pmatrix}

That is, t_i^j=0 for all i\geq j. What happens as we compose this matrix with itself? I say that for T^2 we’ll find the (i,k) entry to be zero for all i\geq {k}+1. Indeed, we can calculate it as a sum of terms like t_i^jtj^k. For each of these factors to be nonzero we need i\leq j-1 and j\leq k-1. That is, i\leq k-2, or else the matrix entry must be zero. Similarly, every additional factor of T pushes the nonzero matrix entries one step further from the diagonal, and eventually they must fall off the upper-right corner. That is, some power of T must give the zero matrix. The vectors may not have been killed by the transformation T, so they may not all have been in the kernel, but they will all be in the kernel of some power of T.

Similarly, let’s take a linear transformation T and a vector v. If v\in\mathrm{Ker}(T-\lambda1_V) we said that v is an eigenvector of T with eigenvalue \lambda. Now we’ll extend this by saying that if v\in\mathrm{Ker}(T-\lambda1_V)^n for some n, then v is a generalized eigenvector of T with eigenvalue \lambda.

About these ads

February 16, 2009 - Posted by | Algebra, Linear Algebra

3 Comments »

  1. […] Generalized Eigenvalues Our definition of a generalized eigenvector looks a lot like the one for an eigenvector. But finding them may not be as straightforward as our […]

    Pingback by Determining Generalized Eigenvalues « The Unapologetic Mathematician | February 17, 2009 | Reply

  2. […] turns out that generalized eigenspaces do capture this notion, and we have a way of calculating them to boot! That is, I’m asserting […]

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

  3. […] Eigenvectors of an Eigenpair Just as we saw when dealing with eigenvalues, eigenvectors alone won’t cut it. We want to consider the […]

    Pingback by Generalized Eigenvectors of an Eigenpair « The Unapologetic Mathematician | April 6, 2009 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: