# The Unapologetic Mathematician

## 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 $\lambda$.

February 16, 2009 Posted by | Algebra, Linear Algebra | 3 Comments