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 $w\ne 0$ (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

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

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 $i=1,\mathrm{\dots },m$. Let $0={\sum }_{i=1}^m{r}_i{v}_i$ with $r_i\in k$. Induct on $i$. If $i=1$, then $v_1=v\ne 0$, 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 $r_1=\mathrm{\cdots }=r_{m-1}=0$ 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 $C_{\lambda }(v_1)\cup \mathrm{\cdots }\cup C_{\lambda }(v_k)$ is linearly independent whenever $\{v_{1\lambda },\mathrm{\dots },v_{k\lambda }\}$ is.

• Let $E=\mathrm{span}(C_{\lambda }(v))$. 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

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

  $T(w_i)=(T-\lambda I+\lambda I){(T-\lambda I)}^{m+1-i}(v)$$=$

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

• 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 $V$.