Generalized Eigenvectors and Eigenspaces

Definition

Suppose $T\in \mathcal{L}(V)$ has an eigenvalue $\lambda$. Its generalized eigenspace is

$$K_{\lambda }:=\left\{\mathbf{x}\in V:(T-\lambda E{)}^k(\mathbf{x})=0\text{ for some }k\in \mathbb{N}\right\}$$

and a generalized eigenvector is any non-zero $\mathbf{v}\in K_{\lambda }$

Relation with characteristic polynomial

Suppose that the characteristic polynomial of $T\in \mathcal{L}(V)$ splits over $F$:

$$\chi (t)=({\lambda }_1-t{)}^{m_1}\cdots ({\lambda }_k-t{)}^{m_k}$$

where the ${\lambda }_j$ are the distinct eigenvalues of $T$ with algebraic multiplicities $m_j$. Then:

1. For each eigenvalue;
   • $K_{\lambda }=\ker (T-\lambda E{)}^m$
   • $\dim K_{\lambda }=m$
2. $V=K_{{\lambda }_1}\oplus \cdots \oplus K_{{\lambda }_k}$, that is, there exists a basis of $V$ consisting generalized eigenvectors

Properties

Let $\lambda$ be an eigenvalue of $T\in \mathcal{L}(V)$. Then

1. The eigenspace $V_{\lambda }$ is a subspace of $K_{\lambda }$
2. $K_{\lambda }$ is $T$-invariant
3. Suppose $K_{\lambda }$ is finite-dimensional and $t\ne \lambda$. Then
   • $K_{\lambda }$ is $(T-kE)$-invariant and the restriction of $T-kE$ to $K_{\lambda }$ is an isomorphism
   • If $t$ is another eigenvalue, then $K_{\lambda }\cap K_t=\left\{0\right\}$.

Basis of a Jordan block

A cycle of generalized eigenvectors for a linear operator $T$ is a set

$${\beta }_{\mathbf{x}}=\left\{(T-\lambda E{)}^{k-1}(\mathbf{x}),\dots ,(T-\lambda E)(\mathbf{x}),\mathbf{x}\right\}$$

where the generator $\mathbf{x}\in K_{\lambda }$ is non-zero and $k$ is minimal such that $(T-\lambda E{)}^k(\mathbf{x})=0$

Then,

1. ${\beta }_{\mathbf{x}}$ is linearly independent and thus a basis of $⟨{\beta }_{\mathbf{x}}⟩$
2. $⟨{\beta }_{\mathbf{x}}⟩$ is $T$-invariant. With respect to ${\beta }_{\mathbf{x}}$, the matrix of the restriction of $T$ is the $k\times k$ Jordan block, that is

$${\left[T_{⟨{\beta }_{\mathbf{x}}⟩}\right]}_{{\beta }_{\mathbf{x}}}=J_k(\lambda )$$