Generalized Eigenvectors

A generalized eigenvector $v$ of $T \in \mathscr{L}(\mathbf{V})$ corresponding to $\lambda \in \sigma (T)$ is a vector satisfying:

$$ (T-\lambda I)^k v = 0 \hspace{25pt} \exists k >0 $$

Lemma. The null space of a power of an operator $T \in \mathscr{L}(\mathbf{V})$ is contained in the null spaces of successive powers, eventually stabilizing.

$$ \text{1. } \textbf {Containment } \hspace{41pt} \{0\} \sub \text{null }T \sub \text{null }T^2 \sub \dots \hspace{64pt}\\

\text{2. } \textbf{Stabilization} \hspace{42pt} \text{null }T^m = \text{null }T^{m+1} \rightarrow \text{null }T^{m+k}=\text{null }T ^m \\

\text{3. } \textbf{Convergence Bound} \hspace{41pt} m\leq \dim \mathbf{V} \hspace{107pt} $$

Proof </aside>

Lemma. Every generalized eigenvector of $T \in \mathscr{L}(\mathbf{V})$ corresponds to exactly one eigenvalue of $T$.

Proof </aside>

Lemma. For distinct eigenvalues $\lambda_1, \dots, \lambda _n$ of $T\in\mathscr{L}(\mathbf{V})$, the list of corresponding generalized eigenvectors $v_1, \dots, v_n$ is linearly independent.

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Theorem. Any vector space $\mathbf{V}$ can be decomposed into the null space and range of any operator $T \in \mathscr{L}(\mathbf{V})$.

$$ \mathbf{V} = \text{null }T^{\dim \mathbf{V}} \oplus \text{range }T^{\dim \mathbf{V}} $$

Proof </aside>

Theorem. For $F = \mathbb{C}$ and $T \in \mathscr{L}(\mathbf{V})$, there exists a basis of $\mathbf{V}$ consisting entirely of generalized eigenvectors of $T$.

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Generalized Eigenspaces

The generalized eigenspace corresponding to an eigenvalue $\lambda$ of $T \in \mathscr{L}(\mathbf{V})$ is the subspace $G(\lambda, T)$ with the following underlying set:

$$ G(\lambda, T) = \{v \in V : (T-\lambda I)^k v = 0 \hspace{10pt} \exists k \in \N \} = \text{null }(T-\lambda I)^{\dim \mathbf{V}} $$

Theorem. For each $\lambda \in \sigma(T)$, its corresponding generalized eigenspace is characterized as follows:

$$ \text{1. }\textbf{Invariance} \hspace{63pt} \text{$G(\lambda, T)$ is $T$-invariant} \\

\text{2. } \textbf{Nilpotence} \hspace{35pt} (T-\lambda I)\restriction_{G(\lambda , T)} \text{ is nilpotent} \\

\text{3. } \textbf{G.E-Space Decomp.} \hspace{32pt} \mathbf{V} = \bigoplus _{\lambda \in \sigma (T)} G(\lambda, T) $$

Proof </aside>

Jordan Form

A basis $\beta$ of $\mathbf{V}$ is Jordan for $T \in \mathscr{L}(\mathbf{V})$ if $[T]_\beta$ is a matrix containing block matrices on the diagonal and zeroes elsewhere.

$$ [T]_\beta = \begin{pmatrix} A_1 & \dots & 0 \\ \vdots & \ddots & \vdots \\

0 & \dots & A_p \end{pmatrix} $$

Lemma. Any nilpotent operator admits a Jordan basis.

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Theorem. For a finite-dimensional $\mathbf{V}$ $\mathbf{V}$$F = \mathbb{C}$ and $T \in \mathscr{L}(\mathbf{V})$, there exists a basis which is Jordan for $T$.

Proof </aside>