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 $w\ne 0$ (since $v\ne 0$ ) 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 length. 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 $0=\sum_{i=1}^m r_iv_i$ with $r_i\in k$ . Induct on $i$ . If $i=1$ , then $v_1=v\ne 0$ , so $r_1=0$ and $\lbrace v_1\rbrace$ is linearly independent. Suppose the property is true when $i=m-1$ . Apply $T-\lambda I$ to the equation, and we have $0=\sum_{i=1}^m r_i(T-\lambda I)(v_i)= \sum_{i=1}^{m-1}r_iv_{i+1}$ . Then $r_1=\cdots=r_{m-1}=0$ by induction. So $0=r_mv_m=r_mv_{\lambda}$ 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 $\lbrace v_{1\lambda},\ldots,v_{k\lambda}\rbrace$ 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
  
  
  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 $V$ .

Bibliography

1

Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.