# generalized eigenvector

Let $V$ be a vector space over a field $k$ and $T$ a linear transformation on $V$ (a linear operator). A non-zero vector $v\in V$ is said to be a generalized eigenvector of $T$ (corresponding to $\lambda$) if there is a $\lambda\in k$ and a positive integer $m$ such that

 $(T-\lambda I)^{m}(v)=0,$

where $I$ is the identity operator.

In the equation above, it is easy to see that $\lambda$ is an eigenvalue of $T$. Suppose that $m$ is the least such integer satisfying the above equation. If $m=1$, then $\lambda$ is an eigenvalue of $T$. If $m>1$, let $w=(T-\lambda I)^{m-1}(v)$. Then $w\neq 0$ (since $v\neq 0$) and $(T-\lambda I)(w)=0$, so $\lambda$ is again an eigenvalue of $T$.

Let $v$ be a generalized eigenvector of $T$ corresponding to the eigenvalue $\lambda$. We can form a sequence

 $v,(T-\lambda I)(v),(T-\lambda I)^{2}(v),\ldots,(T-\lambda I)^{i}(v),\ldots,(T-% \lambda I)^{m}(v)=0,0,\ldots$

The set $C_{\lambda}(v)$ of all non-zero terms in the sequence is called a cycle of generalized eigenvectors of $T$ corresponding to $\lambda$. The cardinality $m$ of $C_{\lambda}(v)$ is its . For any $C_{\lambda}(v)$, write $v_{\lambda}=(T-\lambda I)^{m-1}(v)$.

Below are some properties of $C_{\lambda}(v)$:

• $v_{\lambda}$ is the only eigenvector of $\lambda$ in $C_{\lambda}(v)$, for otherwise $v_{\lambda}=0$.

• $C_{\lambda}(v)$ is linearly independent.

###### Proof.

Let $v_{i}=(T-\lambda I)^{i-1}(v)$, where $i=1,\ldots,m$. Let $0=\sum_{i=1}^{m}r_{i}v_{i}$ with $r_{i}\in k$. Induct on $i$. If $i=1$, then $v_{1}=v\neq 0$, so $r_{1}=0$ and $\{v_{1}\}$ is linearly independent. Suppose the property is true when $i=m-1$. Apply $T-\lambda I$ to the equation, and we have $0=\sum_{i=1}^{m}r_{i}(T-\lambda I)(v_{i})=\sum_{i=1}^{m-1}r_{i}v_{i+1}$. Then $r_{1}=\cdots=r_{m-1}=0$ by induction. So $0=r_{m}v_{m}=r_{m}v_{\lambda}$ and thus $r_{m}=0$ since $v_{\lambda}$ is an eigenvector and is non-zero. ∎

• More generally, it can be shown that $C_{\lambda}(v_{1})\cup\cdots\cup C_{\lambda}(v_{k})$ is linearly independent whenever $\{v_{1\lambda},\ldots,v_{k\lambda}\}$ is.

• Let $E=\operatorname{span}(C_{\lambda}(v))$. Then $E$ is a $(m+1)$-dimensional subspace of the generalized eigenspace of $T$ corresponding to $\lambda$. Furthermore, let $T|_{E}$ be the restriction of $T$ to $E$, then $[T|_{E}]_{C_{\lambda}(v)}$ is a Jordan block, when $C_{\lambda}(v)$ is ordered (as an ordered basis) by setting

 $(T-\lambda I)^{i}(v)<(T-\lambda I)^{j}(v)\qquad\mbox{ whenever }\qquad i>j.$

Indeed, for if we let $w_{i}=(T-\lambda I)^{m+1-i}(v)$ for $i=1,\ldots m+1$, then

 $\displaystyle T(w_{i})=(T-\lambda I+\lambda I)(T-\lambda I)^{m+1-i}(v)$ $\displaystyle=$ $\displaystyle\left\{\begin{array}[]{ll}\lambda w_{i}&\mbox{ if }i=1,\\ w_{i-1}+\lambda w_{i}&\mbox{ otherwise.}\end{array}\right.$

so that $[T|_{E}]_{C_{\lambda}(v)}$ is the $(m+1)\times(m+1)$ matrix given by

 $\begin{pmatrix}\lambda&1&0&\cdots&0\\ 0&\lambda&1&\cdots&0\\ 0&0&\lambda&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ 0&0&0&\cdots&\lambda\end{pmatrix}$
• A cycle of generalized eigenvectors is called maximal if $v\notin(T-\lambda I)(V)$. If $V$ is finite dimensional, any cycle of generalized eigenvectors $C_{\lambda}(v)$ can always be extended to a maximal cycle of generalized eigenvectors $C_{\lambda}(w)$, meaning that $C_{\lambda}(v)\subseteq C_{\lambda}(w)$.

• In particular, any eigenvector $v$ of $T$ can be extended to a maximal cycle of generalized eigenvectors. Any two maximal cycles of generalized eigenvectors extending $v$ span the same subspace of $V$.

## References

• 1 Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.
Title generalized eigenvector GeneralizedEigenvector 2013-03-22 17:23:13 2013-03-22 17:23:13 CWoo (3771) CWoo (3771) 13 CWoo (3771) Definition msc 65F15 msc 65-00 msc 15A18 msc 15-00 GeneralizedEigenspace cycle of generalized eigenvectors