7 Canonical Forms

7.1 The Jordan Canonical Form

From these subspaces, we select ordered bases whose union is an ordered basis $\beta$ for $\mathsf{V}$ such that

$$\begin{equation*} [\mathsf{T}]_\beta = \begin{pmatrix} A_1 & O & \dots & O \\ O & A_2 & \dots & O \\ \vdots & \vdots & & \vdots \\ O & O & \dots & A_k \end{pmatrix}, \end{equation*}$$

where each $O$ is a zero matrix, and each $A_i$ is a square matrix of the form $(\lambda)$ or

$$\begin{equation*} \begin{pmatrix} \lambda & 1 & 0 & \dots & 0 & 0 \\ 0 & \lambda & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \dots & \lambda & 1 \\ 0 & 0 & 0 & \dots & 0 & \lambda \end{pmatrix} \end{equation*}$$

for some eigenvalue $\lambda$ of $\mathsf{T}$. Such a matrix $A_i$ is called a Jordan block corresponding to $\lambda$, and the matrix $[\mathsf{T}]_\beta$ is called a Jordan canonical form of $\mathsf{T}$. We also say that the ordered basis $\beta$ is a Jordan canonical basis for $\mathsf{T}$. Observe that each Jordan block $A_i$ is “almost” a diagonal matrix—in fact, $[\mathsf{T}]_\beta$ is a diagonal matrix if and only if each $A_i$ is of the form $(\lambda)$.

Definition. Let $\mathsf{T}$ be a linear operator on a vector space $\mathsf{V}$, and let $\lambda$ be a scalar. A nonzero vector $x$ in $\mathsf{V}$ is called a generalized eigenvector of $\mathsf{T}$ corresponding to $\lambda$ if

$$\begin{equation*} (\mathsf{T} - \lambda\mathsf{I})^p(x) = 0 \end{equation*}$$

for some positive integer $p$.

Notice that if $x$ is a generalized eigenvector of $\mathsf{T}$ corresponding to $\lambda$, and $p$ is the smallest positive integer for which $(\mathsf{T} - \lambda\mathsf{I})^p(x) = 0$, then $(\mathsf{T} - \lambda\mathsf{I})^{p-1}(x)$ is an eigenvector of $\mathsf{T}$ corresponding to $\lambda$. Therefore $\lambda$ is an eigenvalue of $\mathsf{T}$.

Definition. Let $\mathsf{T}$ be a linear operator on a vector space $\mathsf{V}$, and let $\lambda$ be an eigenvalue of $\mathsf{T}$. The generalized eigenspace of $\mathsf{T}$ corresponding to $\lambda$, denoted $\mathsf{K}_\lambda$, is the subset of $\mathsf{V}$ defined by

$$\begin{equation*} \mathsf{K}_\lambda = \{x \in \mathsf{V} : (\mathsf{T} - \lambda\mathsf{I})^p(x) = 0 \text{ for some positive integer } p\}. \end{equation*}$$

Note that $\mathsf{K}_\lambda$ consists of the zero vector and all generalized eigenvectors corresponding to $\lambda$.

Theorems

1. Let $\mathsf{T}$ be a linear operator on a vector space $\mathsf{V}$, and let $\lambda$ be an eigenvalue of $\mathsf{T}$. Then:

   (a) $\mathsf{K}_\lambda$ is a $\mathsf{T}$-invariant subspace of $\mathsf{V}$ containing $\mathsf{E}_\lambda$.

   (b) For any scalar $\mu \neq \lambda$, the restriction of $\mathsf{T} - \mu\mathsf{I}$ to $\mathsf{K}_\lambda$ is one-to-one.

2. Let $\mathsf{T}$ be a linear operator on a finite-dimensional vector space $\mathsf{V}$ such that the characteristic polynomial of $\mathsf{T}$ splits. Suppose that $\lambda$ is an eigenvalue of $\mathsf{T}$ with multiplicity $m$. Then:

   (a) $\dim(\mathsf{K}_\lambda) \le m$.

   (b) $\mathsf{K}_\lambda = \mathsf{N}((\mathsf{T} - \lambda\mathsf{I})^m)$.

3. Let $\mathsf{T}$ be a linear operator on a finite-dimensional vector space $\mathsf{V}$ such that the characteristic polynomial of $\mathsf{T}$ splits, and let $\lambda_1, \lambda_2, \dots, \lambda_k$ be the distinct eigenvalues of $\mathsf{T}$. Then, for every $x \in \mathsf{V}$, there exist vectors $v_i \in \mathsf{K}_{\lambda_i}$, $1 \le i \le k$, such that

   $$\begin{equation*} x = v_1 + v_2 + \dots + v_k. \end{equation*}$$

4. Let $\mathsf{T}$ be a linear operator on a finite-dimensional vector space $\mathsf{V}$ such that the characteristic polynomial of $\mathsf{T}$ splits, and let $\lambda_1, \lambda_2, \dots, \lambda_k$ be the distinct eigenvalues of $\mathsf{T}$ with corresponding multiplicities $m_1, m_2, \dots, m_k$. For $1 \le i \le k$, let $\beta_i$ be an ordered basis for $\mathsf{K}_{\lambda_i}$. Then the following statements are true.

   (a) $\beta_i \cap \beta_j = \varnothing$ for $i \neq j$.

   (b) $\beta = \beta_1 \cup \beta_2 \cup \dots \cup \beta_k$ is an ordered basis for $\mathsf{V}$.

   (c) $\dim(\mathsf{K}_{\lambda_i}) = m_i$ for all $i$.

Definitions. Let $\mathsf{T}$ be a linear operator on a vector space $\mathsf{V}$, and let $x$ be a generalized eigenvector of $\mathsf{T}$ corresponding to the eigenvalue $\lambda$. Suppose that $p$ is the smallest positive integer for which $(\mathsf{T} - \lambda\mathsf{I})^p(x) = 0$. Then the ordered set

$$\begin{equation*} \{(\mathsf{T} - \lambda\mathsf{I})^{p-1}(x), (\mathsf{T} - \lambda\mathsf{I})^{p-2}(x), \dots, (\mathsf{T} - \lambda\mathsf{I})(x), x\} \end{equation*}$$

is called a cycle of generalized eigenvectors of $\mathsf{T}$ corresponding to $\lambda$. The vectors $(\mathsf{T} - \lambda\mathsf{I})^{p-1}(x)$ and $x$ are called the initial vector and the end vector of the cycle, respectively. We say that the length of the cycle is $p$.

Theorem Let $\mathsf{T}$ be a linear operator on a finite-dimensional vector space $\mathsf{V}$ whose characteristic polynomial splits, and suppose that $\beta$ is a basis for $\mathsf{V}$ such that $\beta$ is a disjoint union of cycles of generalized eigenvectors of $\mathsf{T}$. Then the following statements are true. (a) For each cycle $\gamma$ of generalized eigenvectors contained in $\beta$, $\mathsf{W} = \operatorname{span}(\gamma)$ is $\mathsf{T}$-invariant, and $[\mathsf{T}_{\mathsf{W}}]_\gamma$ is a Jordan block. (b) $\beta$ is a Jordan canonical basis for $\mathsf{V}$.

Theorem. Let $\mathsf{T}$ be a linear operator on a vector space $\mathsf{V}$, and let $\lambda$ be an eigenvalue of $\mathsf{T}$. Suppose that $\gamma_1, \gamma_2, \dots, \gamma_q$ are cycles of generalized eigenvectors of $\mathsf{T}$ corresponding to $\lambda$ such that the initial vectors of the $\gamma_i$'s are distinct and form a linearly independent set. Then the $\gamma_i$'s are disjoint, and their union $\gamma = \bigcup_{i=1}^q \gamma_i$ is linearly independent.