Finished all chapters and definitions. I need to add subsections and see if there's any theorems or definitions in the appendicies that are worth adding to this as well.
This commit is contained in:
@@ -1 +1,132 @@
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\section{The Jordan Canonical Form I}
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\begin{definition}
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In this section, we extend the definition of eigenspace to \textit{generalized eigenspace}. From these subspaces, we select ordered bases whose union is an ordered basis $\beta$ for $V$ such that
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\[[T]_\beta = \begin{pmatrix}
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A_1 & O & \dots & O \\
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O & A_2 & \dots & O \\
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\vdots & \vdots & & \vdots \\
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O & O & \dots & A_k
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\end{pmatrix}\]
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where each $O$ is a zero matrix, and each $A_i$ is a square matrix of the form $(\lambda$) or
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\[\begin{pmatrix}
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\lambda & 1 & 0 & \dots & 0 & 0 \\
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0 & \lambda & 1 & \dots & 0 & 0 \\
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\vdots & \vdots & \vdots & & \vdots & \vdots \\
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0 & 0 & 0 & \dots & \lambda & 1 \\
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0 & 0 & 0 & \dots & 0 & \lambda
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\end{pmatrix}\]
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for some eigenvalue $\lambda$ of $T$. Such a matrix $A_i$ is called a \textbf{Jordan block} corresponding to $\lambda$, and the matrix $[T]_\beta$ is called a \textbf{Jordan canonical form} of $T$. We also say that the ordered basis $\beta$ is a \textbf{Jordan canonical basis} for $T$. Observe that each Jordan block $A_i$ is ``almost" a diagonal matrix -- in fact, $[T]_\beta$ is a diagonal matrix if and only if each $A_i$ is of the form $(\lambda)$.
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\end{definition}
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\begin{definition}
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\hfill\\
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Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a \textbf{generalized eigenvector of $T$ corresponding to $\lambda$} if $(T -\lambda I)^p(x) = 0$ for some positive integer $p$.
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\end{definition}
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\begin{definition}
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\hfill\\
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Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. The \textbf{generalized eigenspace of $T$ corresponding to $\lambda$}, denoted $K_\lambda$, is the subset of $V$ defined by
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\[K_\lambda = \{x \in V : (T - \lambda I)^p(x) = 0\ \text{for some positive integer}\ p\}.\]
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\end{definition}
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\begin{theorem}\label{Theorem 7.1}
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\hfill\\
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Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. then
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\begin{enumerate}
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\item $K_\lambda$ is a $T$-invariant subspace of $V$ containing $E_\lambda$ (the eigenspace of $T$ corresponding to $\lambda$).
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\item for any scalar $\mu \neq \lambda$, the restriction of $T - \mu I$ to $K_\lambda$ is one-to-one.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits. Suppose that $\lambda$ is an eigenvalue of $T$ with multiplicity $m$. Then
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\begin{enumerate}
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\item $\ldim{K_\lambda} \leq m$.
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\item $K_\lambda = \n{(T - \lambda I)^m}$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and let $\lambda_1, \lambda_2, \dots, \lambda_k$ be the distinct eigenvalues of $T$. Then, for every $x \in V$, there exist vectors $v_i \in K_\lambda$, $1 \leq i \leq k$, such that
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\[x = v_1 + v_2 + \dots + v_k.\]
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\end{theorem}
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\begin{theorem}\label{Theorem 7.4}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and let $\lambda_1, \lambda_2, \dots, \lambda_k$ be the distinct eigenvalues of $T$ with corresponding multiplicities $m_1, m_2, \dots, m_k$. For $1 \leq i \leq k$, let $\beta_i$ be an ordered basis for $K_{\lambda_i}$. Then the following statements are true.
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\begin{enumerate}
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\item $\beta_i \cap \beta_j = \emptyset$ for $i \neq j$.
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\item $\beta = \beta_1 \cup \beta_2 \cup \dots \cup \beta_k$ is an ordered basis for $V$.
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\item $\ldim{K_{\lambda_i}} = m_i$ for all $i$.
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\end{enumerate}
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\end{theorem}
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\begin{corollary}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits. Then $T$ is diagonalizable if and only if $E_\lambda = K_\lambda$ for every eigenvalue $\lambda$ of $T$.
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\end{corollary}
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\begin{definition}
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\hfill\\
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Let $T$ be a linear operator on a vector space $V$, and let $x$ be a generalized eigenvector of $T$ corresponding to the eigenvalue $\lambda$. Suppose that $p$ is the smallest positive integer for which $(T - \lambda I)^p(x) = 0$. Then the ordered set
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\[\{(T-\lambda I)^{p-1}(x), (T-\lambda I)^{p -2}(x), \dots, (T-\lambda I)(x), x\}\]
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is called a \textbf{cycle of generalized eigenvectors} of $T$ corresponding to $\lambda$. The vectors $(T-\lambda I)^{p-1}(x)$ and $x$ are called the \textbf{initial vector} and the \textbf{end vector} of the cycle, respectively. We say that the \textbf{length} of the cycle is $p$.
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\end{definition}
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\begin{theorem}\label{Theorem 7.5}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ whose characteristic polynomial splits, and suppose that $\beta$ i a basis for $V$ such that $\beta$ is a disjoint union of cycles of generalized eigenvectors of $T$. Then the following statements are true.
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\begin{enumerate}
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\item For each cycle $\gamma$ of generalized eigenvectors contained in $\beta$, $W = \lspan{\gamma}$ is $T$-invariant, and $[T_W]_\gamma$ is a Jordan block.
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\item $\beta$ is a Jordan canonical basis for $V$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Suppose that $\gamma_1, \gamma_2, \dots,\gamma_q$ are cycles of generalized eigenvectors of $T$ 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 $\gamma = \displaystyle\bigcup_{i = 1}^q \gamma_i$ is linearly independent.
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\end{theorem}
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\begin{corollary}
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\hfill\\
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Every cycle of generalized eigenvectors of a linear operator is linearly independent.
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\end{corollary}
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\begin{theorem}\label{Theorem 7.7}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Then $K_\lambda$ has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to $\lambda$.
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\end{theorem}
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\begin{corollary}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ whose characteristic polynomial splits. then $T$ has Jordan canonical form.
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\end{corollary}
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\begin{definition}
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\hfill\\
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Let $A \in M_{n \times n}(\F)$ be such that the characteristic polynomial of $A$ (and hence of $L_A$) splits. Then the \textbf{Jordan canonical form} of $A$ is defined to be the Jordan canonical form of the linear operator $L_A$ on $\F^n$.
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\end{definition}
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\begin{corollary}
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\hfill\\
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Let $A$ be an $n \times n$ matrix whose characteristic polynomial splits. Then $A$ has Jordan canonical form $J$, and $A$ is similar to $J$.
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\end{corollary}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ whose characteristic polynomial splits. Then $V$ is the direct sum of the generalized eigenspaces of $T$.
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\end{theorem}
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@@ -1 +1,118 @@
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\section{The Jordan Canonical Form II}
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\begin{definition}
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\hfill\\
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For the purposes of this section, we fix a linear operator $T$ on an $n$-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits. Let $\lambda_1, \lambda_2, \dots, \lambda_k$ be the distinct eigenvalues of $T$.\\
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By \autoref{Theorem 7.7}, each generalized eigenspace $K_{\lambda_i}$ contains an ordered basis $\beta_i$ consisting of a union of disjoint cycles of generalized eigenvectors corresponding to $\lambda_i$. So by \autoref{Theorem 7.4}(2) and \autoref{Theorem 7.5}, the union $\beta = \displaystyle\bigcup_{i=1}^k \beta_i$ is a Jordan canonical basis for $T$. For each $i$, let $T_i$ be the restriction of $T$ to $K_{\lambda_i}$, and let $A_i = [T_i]_{\beta_i}$. Then $A_i$ is the Jordan canonical form of $T_{ij}$, and
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\[J = [T]_\beta = \begin{pmatrix}
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A_1 & O & \dots & O \\
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O & A_2 & \dots & O \\
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\vdots & \vdots & & \vdots \\
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O & O & \dots & A_k
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\end{pmatrix}\]
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is the Jordan canonical form of $T$. In this matrix, each $O$ is a zero matrix of appropriate size.\\
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\textbf{Note:} In this section, we compute the matrices $A_i$ and the bases $\beta_i$, thereby computing $J$ and $\beta$ as well. To aid in formulating the uniqueness theorem for $J$, we adopt the following convention: The basis $\beta_i$ for $K_\lambda$ will henceforth be ordered in such a way that the cycles appear in order of decreasing length. That is, if $\beta_i$ is a disjoint union of cycles $\gamma_1, \gamma_2, \dots, \gamma_{n_i}$ and if the length of the cycle $\gamma_j$ is $p_j$, we index the cycles so that $p_1 \geq p_2 \geq \dots \geq p_{n_i}$.\\
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To illustrate the discussion above, suppose that, for some $i$, the ordered basis $\beta_i$ for $K_{\lambda_i}$ is the union of four cycles $\beta_i = \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$ with respective lengths $p_1 = 3, p_2 = 3, p_3 = 2$, and $p_4 = 1$. Then
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\[A_i = \left(\begin{array}{*9{c}}\
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\cellcolor{Gray}\lambda_i & \cellcolor{Gray}1 & \cellcolor{Gray}0 & 0 & 0 & 0 & 0 & 0 & 0 \\
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\cellcolor{Gray}0 & \cellcolor{Gray}\lambda_i & \cellcolor{Gray}1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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\cellcolor{Gray}0 & \cellcolor{Gray}0 & \cellcolor{Gray}\lambda_i & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & \cellcolor{Gray}\lambda_i & \cellcolor{Gray}1 & \cellcolor{Gray}0 & 0 & 0 & 0 \\
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0 & 0 & 0 & \cellcolor{Gray}0 & \cellcolor{Gray}\lambda_i & \cellcolor{Gray}1 & 0 & 0 & 0 \\
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0 & 0 & 0 & \cellcolor{Gray}0 & \cellcolor{Gray}0 & \cellcolor{Gray}\lambda_i & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 & 0 & \cellcolor{Gray}\lambda_i & \cellcolor{Gray}1 & 0 \\
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0 & 0 & 0 & 0 & 0 & 0 & \cellcolor{Gray}0 & \cellcolor{Gray}\lambda_i & 0 \\
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0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cellcolor{Gray}\lambda_i
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\end{array}\right)\]
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To help us visualize each of the matrices $A_i$ and ordered bases $\beta_i$, we use an array of dots called a \textbf{dot diagram} of $T_i$, where $T_i$ is the restriction of $T$ to $K_{\lambda_i}$. Suppose that $\beta_i$ is a disjoint union of cycles of generalized eigenvectors $\gamma_1, \gamma_2, \dots, \gamma_{n_i}$ with lengths $p_1 \geq p_2 \geq \dots \geq p_{n_i}$, respectively. The dot diagram of $T_i$ contains one dot for each vector in $\beta_i$, and the dots are configured according to the following rules.
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\begin{enumerate}
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\item The array consists of $n_i$ columns (one column for each cycle).
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\item Counting from left to right, the $j$th column consists of the $p_j$ dots that correspond to the vectors of $\gamma_j$ starting with the initial vector at the top and continuing down to the end vector.
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\end{enumerate}
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Denote the end vectors of the cycles by $v_1, v_2, \dots, v_{n_i}$. In the following dot diagram of $T_i$, each dot is labeled with the name of the vector in $\beta_i$ to which it corresponds.
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\[\begin{array}{llll}
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\bullet(T - \lambda_i I)^{p_1 - 1}(v_1) & \bullet(T - \lambda_i I)^{p_2-1}(v_2) & \dots & \bullet (T-\lambda_i I)^{p_{n_i} - 1}(v_{n_i}) \\
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\bullet(T - \lambda_i I)^{p_1 - 2}(v_1) & \bullet(T - \lambda_i I)^{p2 - 2}(v_2) & \dots & \bullet(T - \lambda_i I)^{p_{n_i} - 2}(v_{n_i}) \\
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\vdots & \vdots & & \vdots \\
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& & & \bullet(T - \lambda_i I)(v_{n_i}) \\
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& & & \bullet(v_{n_i}) \\
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& \bullet(T - \lambda_i I)(v_2) & & \\
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& \bullet v_2 & & \\
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\bullet(T - \lambda_i I)(v_1) & & \\
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\bullet v_1
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\end{array}\]
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Notice that the dot diagram of $T_i$ has $n_i$ columns (one for each cycle) and $p_1$ rows. Since $p_1 \geq p_2 \geq \dots \geq p_{n_i}$, the columns of the dot diagram become shorter (or at least not longer) as we move from left to right.
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Now let $r_j$ denote the number of dots in the $j$th row of the dot diagram. Observe that $r_1 \geq r_2 \geq \dots \geq r_{p_1}$. Furthermore, the diagram can be reconstructed from the values of the $r_i$'s.\\
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In the above example, with $n_i = 4$, $p_1 = p_2 = 3$, $p_3 = 2$, and $p_4 = 1$, the dot diagram of $T_i$ is as follows:
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\[\begin{array}{llll}
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\bullet & \bullet & \bullet & \bullet \\
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\bullet & \bullet & \bullet & \\
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\bullet & \bullet & &
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\end{array}\]
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Here $r_1 = 4$, $r_2 = 3$ and $r_3 = 2$.
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We now devise a method for computing the dot diagram of $T_i$ using the ranks of linear operators determined by $T$ and $\lambda_i$. Hence the dot diagram is completely determined by $T$, from which it follows that it is unique. On the other hand $\beta_i$ is not unique.
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To determine the dot diagram of $T_i$, we devise a method for computing each $r_j$, the number of dots in the $j$th row of the dot diagram, using only $T_i$ and $\lambda_i$. The next three result give us the required method. To facilitate our arguments, we fix a basis $\beta_i$ for $K_{\lambda_i}$ so that $\beta_i$ is a disjoint union of $n_i$ cycles of generalized eigenvectors with lengths $p_1 \geq p_2 \geq \dots \geq p_{n_i}$.
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\end{definition}
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\begin{theorem}
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\hfill\\
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For any positive integer $r$, the vectors in $\beta_i$ that are associated with the dots in the first $r$ rows of the dot diagram of $T_i$ constitute a basis for $\n{(T - \lambda_i I)^r}$. Hence the number of dots in the first $r$ rows of the dot diagram equals $\nullity{(T - \lambda_i I)^r}$.
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\end{theorem}
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\begin{corollary}
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\hfill\\
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The dimension of $E_{\lambda_i}$ is $n_i$. Hence in a Jordan canonical form of $T$, the number of Jordan blocks corresponding to $\lambda_i$ equals the dimension of $E_{\lambda_i}$.
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\end{corollary}
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\begin{theorem}
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\hfill\\
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Let $r_j$ denote the number of dots in the $j$th row of the dot diagram of $T_i$, the restriction of $T$ to $K_{\lambda_i}$. Then the following statements are true.
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\begin{enumerate}
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\item $r_1 = \ldim{V} - \rank{T - \lambda_i I}$.
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\item $r_j = \rank{(T - \lambda_i I)^{j - 1}} - \rank{(T - \lambda_i I)^j}$ if $j > 1$.
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\end{enumerate}
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\end{theorem}
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\begin{corollary}
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\hfill\\
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For any eigenvalue $\lambda_i$ of $T$, the dot diagram of $T_i$ is unique. Thus, subject to the convention that cycles of generalized eigenvectors for the bases of each generalized eigenspace are listed in order of decreasing length, the Jordan canonical form of a linear operator or a matrix is unique up to the ordering of the eigenvalues.
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\end{corollary}
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\begin{theorem}
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\hfill\\
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Let $A$ and $B$ be $n \times n$ matrices, each having Jordan canonical forms computed according to the conventions of this section. Then $A$ and $B$ are similar if and only if they have (up to an ordering of their eigenvalues) the same Jordan canonical form.
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\end{theorem}
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\begin{lemma}
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A linear operator $T$ on a finite-dimensional vector space $V$ is diagonalizable if and only if its Jordan canonical form is a diagonal matrix. Hence $T$ is diagonalizable if and only if the Jordan canonical basis for $T$ consists of eigenvectors of $T$.
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\end{lemma}
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\begin{definition}
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\hfill\\
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A linear operator $T$ on a vector space $V$ is called \textbf{nilpotent} if $T^p = T_0$ for some positive integer $p$. An $n \times n$ matrix $A$ is called \textbf{nilpotent} if $A^p = O$ for some positive integer $p$.
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\end{definition}
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\begin{definition}
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\hfill\\
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For any $A \in M_{n \times n}(\C)$, define the norm of $A$ by
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\[||A|| = \max \{|A_{ij}| : 1 \leq i, j \leq n\}.\]
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\end{definition}
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@@ -1 +1,70 @@
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\section{The Minimal Polynomial}
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\begin{definition}
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\hfill\\
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A polynomial $f(x)$ with coefficients from a field $\F$ is called \textbf{monic} if its leading coefficient is $1$. If $f(x)$ has positive degree and cannot be expressed as a product of polynomials with coefficients from $\F$ each having positive degree, then $f(x)$ is called \textbf{irreducible}.
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\end{definition}
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\begin{definition}
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Let $T$ be a linear operator on a finite-dimensional vector space. A polynomial $p(t)$ is called a \textbf{minimal polynomial} of $T$ if $p(t)$ is a monic polynomial of least positive degree for which $p(T) = T_0$.
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $p(t)$ be a minimal polynomial of a linear operator $T$ on a finite-dimensional vector space $V$.
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\begin{enumerate}
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\item For any polynomial $g(t)$, if $g(T) = T_0$, then $p(t)$ divides $g(t)$. In particular, $p(t)$ divides the characteristic polynomial of $T$.
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\item The minimal polynomial $T$ is unique.
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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\hfill\\
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Let $A \in M_{n \times n}(\F)$. The \textbf{minimal polynomial} $p(t)$ of $A$ is the monic polynomial of least positive degree for which $p(A) = O$.
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\beta$ be an ordered basis for $V$. Then the minimal polynomial of $T$ is the same as the minimal polynomial of $[T]_\beta$.
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\end{theorem}
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\begin{corollary}
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\hfill\\
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For any $A \in M_{n \times n}(\F)$, the minimal polynomial of $A$ is the same as the minimal polynomial of $L_A$.
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\end{corollary}
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\begin{theorem}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $p(t)$ be the minimal polynomial of $T$. A scalar $\lambda$ is an eigenvalue of $T$ if and only if $p(\lambda) = 0$. Hence the characteristic polynomial and the minimal polynomial of $T$ have the same zeros.
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\end{theorem}
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\begin{corollary}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$ with minimal polynomial $p(t)$ and characteristic polynomial $f(T)$. Suppose that $f(t)$ factors as
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\[f(t) = (\lambda_1 - f)^{n_1}(\lambda_2 - t)^{n_2} \dots(\lambda_k - t)^{n_k},\]
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where $\lambda_1, \lambda_2, \dots, \lambda_k$ are the distinct eigenvalues of $T$. Then there exist integers $m_1, m_2, \dots, m_k$ such that $1 \leq m_i \leq n_i$ for all $i$ and
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\[p(t) = (t - \lambda_1)^{m_1}(t - \lambda_2)^{m_2}\dots(t - \lambda_k)^{m_k}.\]
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\end{corollary}
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\begin{theorem}\label{Theorem 7.15}
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\hfill\\
|
||||
Let $T$ be a linear operator on an $n$-dimensional vector space $V$ such that $V$ is a $T$-cyclic subspace of itself. Then the characteristic polynomial $f(t)$ and the minimal polynomial $p(t)$ have the same degree, and hence $f(t) = (-1)^np(t)$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$. Then $T$ is diagonalizable if and only if the minimal polynomial of $T$ is of the form
|
||||
|
||||
\[p(t) = (t - \lambda_1)(t - \lambda_2) \dots (t - \lambda_k),\]
|
||||
|
||||
where $\lambda_1, \lambda_2, \dots, \lambda_k$ are the distinct eigenvalues of $T$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $x$ be a nonzero vector in $V$. The polynomial $p(t)$ is called a $T$-\textbf{annihilator} of $x$ if $p(t)$ is a monic polynomial of lest degree for which $p(T)(x) = 0$.
|
||||
\end{definition}
|
||||
|
||||
@@ -1 +1,231 @@
|
||||
\section{The rational Canonical Form}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$ with characteristic polynomial
|
||||
|
||||
\[f(t) = (-1)^n(\phi_1(t))^{n_1}(\phi_2(t))^{n_2} \dots (\phi_k(t))^{n_k},\]
|
||||
|
||||
where the $\phi_i(t)$'s ()$1 \leq i \leq k$) are distinct irreducible monic polynomials and the $n_i$'s are positive integers. For $1 \leq i \leq k$, we define the subset $K_{\phi_i}$ of $V$ by
|
||||
|
||||
\[K_{\phi_i} = \{x \in V : (\phi_i(T))^p(x) = 0\ \text{for some positive integer}\ p\}.\]
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $x$ be a nonzero vector in $V$. We use the notation $\mathsf{C}_x$ for the $T$-cyclic subspace generated by $x$. Recall \autoref{Theorem 5.22}, that if $\ldim{\mathsf{C}_x} = k$, then the set
|
||||
|
||||
\[\{x, T(x), T^2(x), \dots, T^{k-1}(x)\}\]
|
||||
|
||||
is an ordered basis for $\mathsf{C}_x$. To distinguish this basis from all other ordered bases for $\mathsf{C}_x$, we call it the $T$-\textbf{cyclic basis generated by \textit{x}} and denote it by $\beta_x$. Let $A$ be the matrix representation of the restriction of $T$ to $\mathsf{C}_x$ relative to the ordered basis $\beta_x$. Recall that
|
||||
|
||||
\[A = \begin{pmatrix}
|
||||
0 & 0 & \dots & 0 & -a_0 \\
|
||||
1 & 0 & \dots & 0 & -a_1 \\
|
||||
0 & 1 & \dots & 0 & -a_2 \\
|
||||
\vdots & \vdots & & \vdots & \vdots \\
|
||||
0 & 0 & \dots & 1 & -a_{k - 1}
|
||||
\end{pmatrix}\]
|
||||
|
||||
where
|
||||
|
||||
\[a_0x + a_1T(x) + \dots + a_{k-1}T^{k-1}(x) + T^k(x) = 0.\]
|
||||
|
||||
furthermore, the characteristic polynomial of $A$ is given by
|
||||
|
||||
\[\det(A - tI) = (-1)^k(a_0 + a_1t + \dots + a_{k-1}t^{k-1} + t^k).\]
|
||||
|
||||
The matrix $A$ is called the \textbf{companion matrix} of the monic polynomial $h(t) = a_0 + a_1t + \dots + a_{k-1}t^{k-1} + t^k$. Every monic polynomial has a companion matrix, and the characteristic polynomial of the companion matrix of a monic polynomial $g(t)$ of degree $k$ is equal to $(-1)^kg(T)$. By \autoref{Theorem 7.15}, the monic polynomial $h(t)$ is also the minimal polynomial of $A$. Since $A$ is the matrix representation of the restriction of $T$ to $\mathsf{C}_x$, $h(t)$ is also the minimal polynomial of this restriction. Note that $h(t)$ is also the $T$-annihilator of $x$.
|
||||
|
||||
It is the object of this section to prove that for every linear operator $T$ on a finite-dimensional vector space $V$, there exists an ordered basis $\beta$ for $V$ such that the matrix representation $[T]_\beta$ is of the form
|
||||
|
||||
\[\begin{pmatrix}
|
||||
C_1 & O & \dots & O \\
|
||||
O & C_2 & \dots & O \\
|
||||
\vdots & \vdots & & \vdots \\
|
||||
O & O & \dots & C_r
|
||||
\end{pmatrix},\]
|
||||
|
||||
where each $C_i$ is the companion matrix of a polynomial $(\phi(t))^m$ such that $phi(t)$ is a monic irreducible divisor of the characteristic polynomial of $T$ and $m$ is a positive integer. A matrix representation of this kind is called a \textbf{rational canonical form} of $T$. We call the accompanying basis a \textbf{rational canonical basis} for $T$.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, let $x$ be a nonzero vector in $V$, and suppose that the $T$-annihilator of $x$ is of the form $(\phi(t))^p$ for some irreducible monic polynomial $\phi(t)$. Then $\phi(t)$ divides the minimal polynomial of $T$, and $x \in K_\phi$.
|
||||
\end{lemma}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\beta$ be an ordered basis for $V$. Then $\beta$ is a rational canonical basis for $T$ if and only if $\beta$ is the disjoint union of $T$-cyclic bases $\beta_{v_i}$, where each $v_i$ lies in $K_\phi$ for some irreducible monic divisor $\phi(t)$ of the characteristic polynomial of $T$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and suppose that
|
||||
|
||||
\[p(t) = (\phi_1(t))^{m_1} (\phi_2(t))^{m_2} \dots (\phi_k(t))^{m_k}\]
|
||||
|
||||
is the minimal polynomial of $T$, where the $\phi_i(t)$'s ($1 \leq i \leq k$) are the distinct irreducible monic factors of $p(t)$ and the $m_i$'s are the positive integers. Then the following statements are true.
|
||||
|
||||
\begin{enumerate}
|
||||
\item $K_{\phi_i}$ is a nonzero $T$-invariant subspace of $V$ for each $i$.
|
||||
\item If $x$ is a nonzero vector in some $K_{\phi_i}$, then the $T$-annihilator of $x$ is of the form $(\phi_i(t))^p$ for some integer $p$.
|
||||
\item $K_{\phi_i} \cap K_{\phi_j} = \{0\}$ for $i \neq j$.
|
||||
\item $K_{\phi_i}$ is invariant under $\phi_j(T)$ for $i \neq j$, and the restriction of $\phi_j(T)$ to $K_{\phi_i}$ is one-to-one and onto.
|
||||
\item $K_{\phi_i} = \n{(\phi_i(T))^{m_i}}$ for each $i$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{lemma}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and suppose that
|
||||
|
||||
\[p(t) = (\phi_1(t))^{m_1} (\phi_2(t))^{m_2} \dots (\phi_k(t))^{m_k}\]
|
||||
|
||||
is the minimal polynomial of $T$, where the $\phi_i$'s ($1 \leq i \leq k$) are the distinct irreducible monic factors of $p(t)$ and the $m_i$'s are the positive integers. For $1 \leq i \leq k$, let $v_i \in K_{\phi_i}$ be such that
|
||||
|
||||
\[v_1 + v_2 + \dots + v_k = 0.\]
|
||||
|
||||
Then $v_i = 0$ for all $i$.
|
||||
\end{lemma}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and suppose that
|
||||
|
||||
\[p(t) = (\phi_1(t))^{m_1} (\phi_2(t))^{m_2} \dots (\phi_k(t))^{m_k}\]
|
||||
|
||||
is the minimal polynomial of $T$, where the $\phi_i$'s ($1 \leq i \leq k$) are the distinct irreducible monic factors of $p(t)$ and the $m_i$'s are the positive integers. For $1 \leq i \leq k$, let $S_i$ be a linearly independent subset of $K_{\phi_i}$. Then
|
||||
|
||||
\begin{enumerate}
|
||||
\item $S_i \cap S_j = \emptyset$ for $i \neq j$.
|
||||
\item $S_1 \cup S_2 \cup \dots \cup S_k$ is linearly independent.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $v_1, v_2, \dots, v_k$ be distinct vectors in $K_\phi$ such that
|
||||
|
||||
\[S_1 = \beta_{v_1} \cup \beta_{v_2} \cup \dots \cup \beta_{v_k}\]
|
||||
|
||||
is linearly independent. For each $i$, choose $w_i \in V$ such that $\phi(T)(w_i) = v_i$. Then
|
||||
|
||||
\[S_2 = \beta_{w_1} \cup \beta_{w_2} \cup \dots \cup \beta_{w_k}\]
|
||||
|
||||
is also linearly independent.
|
||||
\end{theorem}
|
||||
|
||||
\begin{lemma}
|
||||
\hfill\\
|
||||
Let $W$ be a $T$-invariant subspace of $K_\phi$, and let $\beta$ be a basis for $W$. Then the following statements are true.
|
||||
|
||||
\begin{enumerate}
|
||||
\item Suppose that $x \in \n{\phi(T)}$, but $x \notin W$. Then $\beta \cup \beta_x$ is linearly independent.
|
||||
\item For some $w_1, w_2, \dots, w_s$ in $\n{\phi(T)}$, $\beta$ can be extended to the linearly independent set
|
||||
|
||||
\[\beta' = \beta \cup \beta_{w_1} \cup \beta_{w_2} \cup \dots \cup \beta_{w_s},\]
|
||||
|
||||
whose span contains $\n{\phi(T)}$.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
If the minimal polynomial of $T$ is of the form $p(t) = (\phi(t))^m$, then there exists a rational canonical basis for $T$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
$K_\phi$ has a basis consisting of the union of $T$-cyclic bases.
|
||||
\end{corollary}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Every linear operator on a finite-dimensional vector space has a rational canonical basis and, hence, a rational canonical form.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on an $n$-dimensional vector space $V$ with characteristic polynomial
|
||||
|
||||
\[f(t) = (-1)^n(\phi_1(t))^{n_1} (\phi_2(t))^{n_2} \dots (\phi_k(t))^{n_k},\]
|
||||
|
||||
where the $\phi_i(t)$'s ($1 \leq i \leq k$) are distinct irreducible monic polynomials and the $n_i$'s are positive integers. Then the following statements are true.
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\phi_1(t), \phi_2(t), \dots, \phi_k(t)$ are the irreducible monic factors of the minimal polynomial.
|
||||
\item For each $i$, $\ldim{K_{\phi_i}} = d_in_i$, where $d_i$ is the degree of $\phi_i(t)$.
|
||||
\item If $\beta$ is a rational canonical basis for $T$, then $\beta_i = \beta \cap K_{\phi_i}$ is a basis for $K_{\phi_i}$ for each $i$.
|
||||
\item If $\gamma_i$ is a basis for $K_{\phi_i}$ for each $i$, then $\gamma = \gamma_1 \cup \gamma_2 \cup \dots \cup \gamma_k$ is a basis for $V$. In particular, if each $\gamma_i$ is a disjoint union of $T$-cyclic bases, then $\gamma$ is a rational canonical basis for $T$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $\beta$ be a rational canonical basis for $T$, and $\beta_{v_1}, \beta_{v_2}, \dots, \beta_{v_k}$ be the $T$-cyclic bases of $\beta$ that are contained in $K_\phi$. Consider these $T$-cyclic bases $\beta_{v_i}$, and suppose again that the $T$-annihilator of $v_j$ is $(\phi(t))^{p_j}$. Then $\beta_{v_j}$ consists of $dp_j$ vectors in $\beta$, where $d$ is the degree of the polynomial. For $0 \leq i < d$, let $\gamma_i$ be the cycle of generalized eigenvectors of $U$ corresponding to $\lambda = 0$ with end vector $T^i(v_j)$, where $T^0(v_j) = b_j$. Then
|
||||
|
||||
\[\gamma_i = \{(\phi(T))^{p_j-1}T^i(v_j), (\phi(T))^{p_j-2}T^i(v_j), \dots, (\phi(T))T^i(v_j),T^i(v_j)\}.\]
|
||||
|
||||
By \autoref{Theorem 7.1}, $\gamma_i$ is a linearly independent subset of $\mathsf{C}_{v_i}$. Now let
|
||||
|
||||
\[\alpha_j = \gamma_0 \cup \gamma_1 \cup \dots \cup \gamma_{d - 1}.\]
|
||||
|
||||
Notice that $\alpha_j$ contains $p_jd$ vectors.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\hfill\\
|
||||
$\alpha_j$ is an ordered basis for $\mathsf{C}_{v_j}$.
|
||||
\end{lemma}
|
||||
|
||||
\begin{lemma}
|
||||
\hfill\\
|
||||
$\alpha$ is a Jordan canonical basis for $K_\phi$.
|
||||
\end{lemma}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on a finite-dimensional vector space $V$, let $\phi(t)$ be an irreducible monic divisor of the characteristic polynomial of $T$ of degree $d$, and let $r_i$ denote the number of dots in the $i$th row of the dot diagram for $\phi(t)$ with respect to a rational canonical basis for $T$. Then
|
||||
|
||||
\begin{enumerate}
|
||||
\item $r_1 = \displaystyle\frac{1}{d}[\ldim{V}-\rank{\phi(T)}]$
|
||||
\item $r_i = \displaystyle\frac{1}{d}[\rank{(\phi(T))^{i-1}} - \rank{(\phi(t))^i}]$ for $i > 1$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Under the conventions described earlier, the rational canonical form of a linear operator is unique up to the arrangement of the irreducible monic divisors of the characteristic polynomial.
|
||||
\end{corollary}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Since the rational canonical form of a linear operator is unique, the polynomials corresponding to the companion matrices that determine this form are also unique. These polynomials, which are powers of the irreducible monic divisors, are called the \textbf{elementary divisors} of the linear operator. Since a companion matrix may occur more than once in a rational canonical form, the same is true for the elementary divisors. We call the number of such occurrences the \textbf{multiplicity} of the elementary divisor.
|
||||
|
||||
Conversely, the elementary divisors and their multiplicities determine the companion matrices and, therefore, the rational canonical form of a linear operator.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\F)$. The \textbf{rational canonical form} of $A$ is defined to be the rational canonical form of $L_A$. Likewise, for $A$, the \textbf{elementary divisors} and their \textbf{multiplicities} are the same as those of $L_A$.
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[\textbf{Primary Decomposition Theorem}]
|
||||
\hfill\\
|
||||
Let $T$ be a linear operator on an $n$-dimensional vector space $V$ with characteristic polynomial
|
||||
|
||||
\[f(t) = (-1)^n(\phi_1(t))^{n_1} (\phi_2(t))^{n_2} \dots (\phi_k(t))^{n_k},\]
|
||||
|
||||
where the $\phi_i(t)$'s ($1 \leq i \leq k$) are distinct irreducible monic polynomials and the $n_i$'s are positive integers. Then the following statements are true.
|
||||
|
||||
\begin{enumerate}
|
||||
\item $V = K_{\phi_i} \oplus K_{\phi_2} \oplus \dots \oplus K_{\phi_k}$.
|
||||
\item If $T_i$ ($1 \leq i \leq k$) is the restriction of $T$ to $K_{\phi_i}$ and $C_i$ is the rational canonical form of $T_i$, then $C_1 \oplus C_2 \oplus \dots \oplus C_k$ is the rational canonical form of $T$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let$T$ be a linear operator on a finite-dimensional vector space $V$. Then $V$ is a direct sum of $T$-cyclic subspaces $\mathsf{C}_{v_i}$, where each $v_i$ lies in $K_\phi$ for some irreducible monic divisor $\phi(t)$ of the characteristic polynomial of $T$.
|
||||
\end{theorem}
|
||||
|
||||
Reference in New Issue
Block a user