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 +1,117 @@
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\section{Diagonalizability}
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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 $\lambda_1, \lambda_2, \dots, \lambda_k$ be distinct eigenvectors of $T$. If $v_1, v_2, \dots, v_k$ are eigenvectors of $T$ such that $\lambda_i$ corresponds to $v_i$ ($1 \leq i \leq k$), then $\{v_1, v_2, \dots, v_k\}$ is linearly independent.
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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 an $n$-dimensional vector space $V$. If $T$ has $n$ distinct eigenvalues, then $T$ is diagonalizable.
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\end{corollary}
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\begin{definition}
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\hfill\\
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A polynomial $f(t)$ in $P(\F)$ \textbf{splits over} $\F$ if there are scalars $c, a_1, \dots, a_n$ (not necessarily distinct) in $\F$ such that
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\[f(t) = c(t-a_1)(t-a_2)\dots(t-a_n).\]
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\end{definition}
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\begin{theorem}
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\hfill\\
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The characteristic polynomial of any diagonalizable linear operator splits.
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\end{theorem}
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\begin{definition}
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\hfill\\
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Let $\lambda$ be an eigenvalue of a linear operator or matrix with characteristic polynomial $f(t)$. The \textbf{(algebraic) multiplicity} of $\lambda$ is the largest positive integer $k$ for which $(t - \lambda)^k$ is a factor of $f(t)$.
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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 $T$. Define $E_\lambda = \{x \in V : T(x) = \lambda x \}=\n{T - \lambda I_V}$. The set $E_\lambda$ is called the \textbf{eigenspace} of $T$ corresponding to the eigenvalue $\lambda$. Analogously, we define the \textbf{eigenspace} of a square matrix $A$ to be the eigenspace of $L_A$.
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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 $\lambda$ be an eigenvalue of $T$ having multiplicity $m$. Then $1 \leq \ldim{E_\lambda} \leq m$.
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\end{theorem}
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\begin{lemma}
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\hfill\\
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Let $T$ be a linear operator, and let $\lambda_1, \lambda_2, \dots, \lambda_k$ be distinct eigenvalues of $T$. For each $i=1, 2, \dots, k$, let $v_i \in E_{\lambda_i}$, the eigenspace corresponding to $\lambda_i$. If
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\[v_1 + v_2 + \dots + v_k = 0,\]
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then $v_i = 0$ for all $i$.
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\end{lemma}
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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 $\lambda_1, \lambda_2, \dots, \lambda_k$ be distinct eigenvalues of $T$. For each $i = 1, 2, \dots, k$, let $S_i$ be a finite linearly independent subset of the eigenspace $E_{\lambda_i}$. Then $S = S_1 \cup S_2 \cup \dots \cup S_k$ is a linearly independent subset of $V$.
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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. Let $\lambda_1, \lambda_2, \dots \lambda_k$ be the distinct eigenvalues of $T$. Then
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\begin{enumerate}
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\item $T$ is diagonalizable if and only if the multiplicity of $\lambda_i$ is equal to $\ldim{E_{\lambda_i}}$ for all $i$.
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\item If $T$ is diagonalizable and $\beta_i$ is an ordered basis for $E_{\lambda_i}$ for each $i$, then $\beta = \beta_1 \cup \beta_2 \cup \dots \cup \beta_k$ for an ordered basis for $V$ consisting of eigenvectors of $T$.\\
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\textbf{Note:} We regard $\beta_1 \cup \beta_2 \cup \dots \cup \beta_k$ as an ordered basis in the natural way -- the vectors in $\beta_1$ are listed first (in the same order as in $\beta_1$), then the vectors in $\beta_2$ (in the same order as $\beta_2$), etc.
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\end{enumerate}
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\end{theorem}
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\begin{remark}[\textbf{Test for Diagonalization}]
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\hfill\\
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Let $T$ be a linear operator on an $n$-dimensional vector space $V$. Then $T$ is diagonalizable if and only if both of the following conditions hold.
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\begin{enumerate}
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\item The characteristic polynomial of $T$ splits.
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\item For each eigenvalue $\lambda$ of $T$, the multiplicity of $\lambda$ equals $n - \rank{T - \lambda I}$.
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\end{enumerate}
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\end{remark}
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\begin{definition}
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\hfill\\
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Let $W_1, W_2, \dots, W_k$ be subspaces of a vector space $V$. We define the \textbf{sum} of these subspaces to be the set
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\[\{v_1 + v_2 + \dots + v_k : v_i \in W_i\ \text{for}\ 1 \leq i \leq k\}\]
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which we denote by $W_1 + W_2 + \dots + W_k$ or $\displaystyle\sum_{i=1}^{k}W_i$.
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\end{definition}
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\begin{definition}
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\hfill\\
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Let $W_1, W_2, \dots, W_k$ be subspaces of a vector space $V$. We call $V$ the \textbf{direct sum} of the subspaces $W_1, W_2, \dots, W_k$ and write $V = W_1 \oplus W_2 \oplus \dots \oplus W_k$, if
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\[V = \sum_{i=1}^{k}W_i\]
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and
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\[W_j \cap \sum_{i \neq j} W_i = \{0\}\ \ \ \text{for each}\ j\ (1 \leq j \leq k)\]
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $W_1, W_2, \dots, W_k$ be subspaces of a finite-dimensional vector space $V$. The following conditions are equivalent.
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\begin{enumerate}
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\item $V = W_1 \oplus W_2 \oplus \dots \oplus W_k$.
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\item $V = \displaystyle\sum_{i=1}^{k}W_i$ and, for any vectors $v_1, v_2, \dots, v_k$ such that $v_i \in W_i$ ($1 \leq i \leq k$), if $v_1 + v_2 + \dots + v_k = 0$, then $v_i = 0$ for all $i$.
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\item Each vector $v \in V$ can be uniquely written as $v = v_1 + v_2 + \dots + v_k$, where $v_i \in W_i$.
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\item If $\gamma_i$ is an ordered basis for $W_i$ ($1 \leq i \leq k$), then $\gamma_1 \cup \gamma_2 \cup \dots \cup \gamma_k$ is an ordered basis for $V$.
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\item For each $i = 1, 2, \dots, k$, there exists an ordered basis $\gamma_i$ for $W_i$ such that $\gamma_i \cup \gamma_2 \cup \dots \cup \gamma_k$ is an ordered 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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A linear operator $T$ on a finite-dimensional vector space $V$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$.
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\end{theorem}
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\begin{definition}
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\hfill\\
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Two linear operators $T$ and $U$ on a finite-dimensional vector space $V$ are called \textbf{simultaneously diagonalizable} if there exists an ordered basis $\beta$ or $V$ such that both $[T]_\beta$ and $[U]_\beta$ are diagonal matrices. Similarly, $A, B \in M_{n \times n}(\F)$ are called \textbf{simultaneously diagonalizable} if there exists an invertible matrix $Q \in M_{n \times n}(\F)$ such that both $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal matrices.
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\end{definition}
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@@ -1 +1,64 @@
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\section{Eigenvalues and Eigenvectors}
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\begin{definition}
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\hfill\\
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A linear operator $T$ on a finite-dimensional vector space $V$ is called \textbf{diagonalizable} if there is an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix. A square matrix $A$ is called \textbf{diagonalizable} if $L_A$ is diagonalizable.
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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 $V$. A nonzero vector $v \in V$ is called an \textbf{eigenvector} of $T$ if there exists a scalar $\lambda$ such that $T(v) = \lambda v$. The scalar $\lambda$ is called the \textbf{eigenvalue} corresponding to the eigenvector $v$.
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Let $A$ be in $M_{n \times n}(\F)$. A nonzero vector $v \in \F^n$ is called an \textbf{eigenvector} of $A$ if $v$ is an eigenvector of $L_A$; that is, if $Av = \lambda v$ for some scalar $\lambda$. The scalar $\lambda$ is called the \textbf{eigenvalue} of $A$ corresponding to the eigenvector $v$.\\
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The words \textit{characteristic vector} and \textit{proper vector} are also used in place of \textit{eigenvector}. The corresponding terms for \textit{eigenvalue} are \textit{characteristic value} and \textit{proper value}.
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\end{definition}
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\begin{theorem}
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\hfill\\
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A linear operator $T$ on a finite-dimensional vector space $V$ is diagonalizable if and only if there exists an ordered basis $\beta$ for $V$ consisting of eigenvectors of $T$. Furthermore, if $T$ is diagonalizable, $\beta = \{v_1, v_2, \dots, v_n\}$ is an ordered basis of eigenvectors of $T$, and $D=[T]_\beta$, then $D$ is a diagonal matrix and $D_{ij}$ is the eigenvalue corresponding to $v_j$ for $1 \leq j \leq n$.
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\end{theorem}
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\begin{remark}
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\hfill\\
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To \textit{diagonalize} a matrix or linear operator is to find a basis of eigenvectors and the corresponding eigenvalues.
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\end{remark}
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\begin{theorem}
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\hfill\\
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Let $A \in M_{n \times n}(\F)$. Then a scalar $\lambda$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I_n) = 0$.
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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 polynomial $f(t) = \det(A - tI_n)$ is called the \textbf{characteristic polynomial} of $A$.\\
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The observant reader may have noticed that the entries of the matrix $A - tI_n$ are not scalars in the field $\F$. They are, however, scalars in another field $\F(t)$, the field of quotients of polynomials in $t$ with coefficients from $\F$. Consequently, any results proved about determinants in \autoref{Chapter 4} remain valid in this context.
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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 an $n$-dimensional vector space $V$ with ordered basis $\beta$. We define the \textbf{characteristic polynomial} $f(t)$ of $T$ to be the characteristic polynomial of $A=[T]_\beta$. That is,
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\[f(t) = \det(A - tI_n).\]
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $A \in M_{n \times n}(\F)$.
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\begin{enumerate}
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\item The characteristic polynomial of $A$ is a polynomial of degree $n$ with leading coefficient $(-1)^n$.
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\item $A$ has at most $n$ distinct eigenvalues.
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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 $T$. A vector $v$ is an eigenvector of $T$ corresponding to $\lambda$ if and only if $v \neq 0$ and $v \in \n{T - \lambda I)}$.
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\end{theorem}
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\begin{definition}
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\hfill\\
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A \textbf{scalar matrix} is a square matrix of the form $\lambda I$ for some scalar $\lambda$; that is, a scalar matrix is a diagonal matrix in which all the diagonal entries are equal.
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\end{definition}
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@@ -1 +1,81 @@
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\section{Invariant Subspaces and the Cayley-Hamilton Theorem}
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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 $V$. A subspace $W$ of $V$ is called a \textbf{$T$-invariant subspace} of $V$ if $T(W) \subseteq W$, that is, if $T(v) \in W$ for all $v \in W$.
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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 $x$ be a nonzero vector in $V$. The subspace
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\[W = \lspan{\{x, T(x), T^2(x), \dots\}}\]
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is called the \textbf{$T$-cyclic subspace of $V$ generated by $x$}.
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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 $W$ be a $T$-invariant subspace of $V$. Then the characteristic polynomial of $T_W$ divides the characteristic polynomial of $T$.
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\end{theorem}
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\begin{theorem}\label{Theorem 5.22}
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $W$ denote the $T$-cyclic subspace of $V$ generated by a nonzero vector $v \in V$. Let $k = \ldim(W)$. Then
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\begin{enumerate}
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\item $\{v, T(v), T^2(v), \dots, T^{k-1}(v)\}$ is a basis for $W$.
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\item If $a_0v + a_1T(v) + \dots + a_{k-1}T^{k-1}(v)+T^k(v) = 0$, then the characteristic polynomial of $T_W$ is $f(t) = (-1)^k(a_o + a_1t + \dots +a_{k-1}t^{k-1}+t^k)$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{Cayley-Hamilton}]
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\hfill\\
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Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(T) = T_0$, the zero transformation. That is, $T$ ``satisfies" its characteristic equation.
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\end{theorem}
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\begin{corollary}[\textbf{Cayley-Hamilton Theorem for Matrices}]
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\hfill\\
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Let $A$ be an $n \times n$ matrix, and let $f(t)$ be the characteristic polynomial of $A$. Then $f(A) = O$, the $n \times n$ zero matrix.
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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 suppose that $V = W_1 \oplus W_2 \oplus \dots \oplus W_k$, where $W_i$ is a $T$-invariant subspace of $V$ for each $i$ ($1 \leq i \leq k$). Suppose that $f_i(t)$ is the characteristic polynomial of $T_{W_i}$ ($1 \leq i \leq k$). Then $f_1(t)\cdot f_2(t) \cdot \dots \cdot f_k(t)$ is the characteristic polynomial of $T$.
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\end{theorem}
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\begin{definition}
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\hfill\\
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Let $B_1 \in M_{m \times m}(\F)$, and let $B_2 \in M_{n \times n}(\F)$. We define the \textbf{direct sum} of $B_1$ and $B_2$, denoted $B_1 \oplus B_2$, as the $(m + n) \times (m + n)$ matrix $A$ such that
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\[A_{ij} = \begin{cases}
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(B_1)_{ij} & \text{for}\ 1 \leq i, j \leq m \\
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(B_2)_{(i-m),(j-m)} & \text{for}\ m + 1 \leq i, j \leq n + m \\
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0 & \text{otherwise.}
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\end{cases}\]
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If $B_1, B_2, \dots, B_k$ are square matrices with entries from $\F$, then we define the \textbf{direct sum} of $B_1, B_2, \dots, B_k$ recursively by
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\[B_1 \oplus B_2 \oplus \dots \oplus B_k = (B_1 \oplus B_2 \oplus \dots \oplus B_{k-1})\oplus B_k.\]
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If $A= B_1 \oplus B_2 \oplus \dots \oplus B_k$, then we often write
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\[A = \begin{pmatrix}
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B_1 & O & \dots & O \\
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O & B_2 & \dots & O \\
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\vdots & \vdots & & \vdots \\
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O & O & \dots & B_k
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\end{pmatrix}\]
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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 $W_1, W_2, \dots, W_k$ be $T$-invariant subspaces of $V$ such that $V = W_1 \oplus W_2 \oplus \dots \oplus W_k$. For each $i$, let $\beta_i$ be an ordered basis for $W_i$, and let $\beta = \beta_1 \cup \beta_2 \cup \dots \cup \beta_k$. Let $A = [T]_\beta$ and $B_i = [T_{W_i}]_{\beta_i}$ for $i = 1, 2, \dots, k$. Then $A = B_1 \oplus B_2 \oplus \dots \oplus B_k$.
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\end{theorem}
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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 $W$ be a $T$-invariant subspace of $V$. Define $\overline{T}: V/W \to V/W$ by
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\[\overline{T}(v + W) = T(v) + W\ \ \ \text{for any}\ v + W \in V/W.\]
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\end{definition}
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@@ -1 +1,223 @@
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\section{Matrix Limits and Markov Chains}
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\begin{definition}
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\hfill\\
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Let $L, A_1, A_2, \dots$ be $n \times p$ matrices having complex entries. The sequence $A_1, A_2, \dots$ is said to \textbf{converge} to the $n \times p$ matrix $L$, called the \textbf{limit} of the sequence, if
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\[\lim_{m \to \infty}(A_m)_{ij} = L_{ij}\]
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for all $1 \leq i \leq n$ and $1 \leq j \leq p$. To designate that $L$ is the limit of the sequence, we write
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\[\lim_{m \to \infty}A_m = L.\]
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $A_1, A_2, \dots$ be a sequence of $n \times p$ matrices with complex entries that converges to the matrix $L$. Then for any $P \in M_{r \times n}(\C)$ and $Q \in M_{p \times s}(\C)$,
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\[\lim_{m \to \infty}PA_m = PL\ \ \ \ \text{and}\ \ \ \ \lim_{m \to \infty}A_mQ = LQ.\]
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\end{theorem}
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\begin{corollary}
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\hfill\\
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Let $A \in M_{n \times n}(\C)$ be such that $\displaystyle\lim_{m \to \infty}A^m = L$. Then for any invertible matrix $Q \in M_{n \times n}(\C)$,
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\[\lim_{m \to \infty}(QAQ^{-1})^m = QLQ^{-1}.\]
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\end{corollary}
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\begin{theorem}
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\hfill\\
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Let $A$ be a square matrix with complex entries. Then $\displaystyle\lim_{m \to \infty}A^m$ exists if and only if both of the following conditions hold.
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\begin{enumerate}
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\item Every eigenvalue of $A$ is contained in $S$.
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\item If $1$ is an eigenvalue of $A$, then the dimension of the eigenspace corresponding to $1$ equals the multiplicity of $1$ as an eigenvalue of $A$.
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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 $A \in M_{n \times n}(\C)$ satisfy the following two conditions.
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\begin{enumerate}
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\item Every eigenvalue of $A$ is contained in $S$.
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\item $A$ is diagonalizable.
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\end{enumerate}
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Then $\displaystyle\lim_{m \to \infty}A^m$ exists.
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\end{theorem}
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\begin{definition}
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Any square matrix having the following two properties is called a \textbf{transition matrix} or a \textbf{stochastic matrix}.
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\begin{enumerate}
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\item All entries are non-negative
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\item The sum of entries in each column sums up to $1$.
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\end{enumerate}
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For an arbitrary $n \times n$ transition matrix $M$, the rows and columns correspond to $n$ \textbf{states}, and the entry $M_{ij}$ represents the probability of moving from state $j$ to state $i$ in one \textbf{stage}.
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\end{definition}
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\begin{definition}
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A vector with non-negative entries that add up to $1$ is called a \textbf{probability vector}.
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\end{definition}
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||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $M$ be an $n \times n$ matrix having real non-negative entries, let $v$ be a column vector in $\R^n$ having non-negative coordinates, and let $u \in \R^n$ be the column vector in which each coordinate equals $1$. Then
|
||||
|
||||
\begin{enumerate}
|
||||
\item $M$ is a transition matrix if and only if $M^tu = u$;
|
||||
\item $v$ is a probability vector if and only if $u^tv = (1)$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\begin{enumerate}
|
||||
\item[]
|
||||
\item The product of two $n \times n$ transition matrices is an $n \times n$ transition matrix. In particular, any power of a transition matrix is a transition matrix.
|
||||
\item The product of a transition matrix and a probability vector is a probability vector.
|
||||
\end{enumerate}
|
||||
\end{corollary}
|
||||
|
||||
\begin{definition}
|
||||
A process in which elements of a set are each classified as being in one of several fixed states that can switch over time is called a \textbf{stochastic process}. The switching to a particular state is described by the probability, and in general this probability depends on such factors as the state in question, the time in question, some or all of the previous states in which the object has been (including the current state), and the states that other objects are in or have been in.\\
|
||||
|
||||
If, however, the probability that an object in one state changes to a different state in a fixed interval of time depends only on the two states (and not on the time, earlier states, or other factors), then the stochastic process is called a \textbf{Markov process}. If, in addition, the number of possible states is finite, then the Markov process is called a \textbf{Markov chain}.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
The vector that describes the initial probability of being in each state of a Markov chain is called the \textbf{initial probability vector} for the Markov chain.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
A transition matrix is called \textbf{regular} if some power of the matrix contains only positive entries.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\C)$. For $1 \leq i, j \leq n$, define $\rho_i(A)$ to be the sum of the absolute values of the entries of row $i$ of $A$, and define $\nu_j(A)$ to be equal to the sum of the absolute values of the entries of column $j$ and $A$. Thus
|
||||
|
||||
\[\rho_i(A) = \sum_{j=1}^{n}|A_{ij}|\ \ \text{for}\ i=1, 2, \dots, n\]
|
||||
|
||||
and
|
||||
|
||||
\[\nu_j(A) = \sum_{i=1}^{n}|A_{ij}|\ \ \text{for}\ j=1, 2, \dots n.\]
|
||||
|
||||
The \textbf{row sum} of $A$, denoted $\rho(A)$, and the \textbf{column sum} of $A$, denoted $\nu(A)$, are defined as
|
||||
|
||||
\[\rho(A) = \max\{\rho_i(A) : 1 \leq i \leq n\}\ \ \ \ \text{and}\ \ \ \ \nu(A) = \max\{\nu_j(A) : 1 \leq j \leq n\}.\]
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
For an $n \times n$ matrix $A$, we define the $i$th \textbf{Gerschgorin disk} $\C_i$ to be the disk in the complex plane with center $A_{ij}$ and radius $r_i = \rho_i(A) - |A_{ii}|$; that is,
|
||||
|
||||
\[\C_i = \{z \in \C:|z - A_{ii}| < r_i\}\]
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[\textbf{Gerschgorin's Disk Theorem}]
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\C)$. Then ever eigenvalue of $A$ is contained in a Gerschgorin disk.
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Let $\lambda$ be any eigenvalue of $A \in M_{n \times n}(\C)$. Then $|\lambda| \leq \rho(A)$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Let $\lambda$ be any eigenvalue of $A \in M_{n \times n}(\C)$. Then
|
||||
|
||||
\[|\lambda| \leq \min\{\rho(A), \nu(A)\}.\]
|
||||
\end{corollary}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
If $\lambda$ is an eigenvalue of a transition matrix, then $|\lambda| \leq 1$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Every transition matrix has $1$ as an eigenvalue.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\C)$ be a matrix in which each entry is positive, and let $\lambda$ be an eigenvalue of $A$ such that $|\lambda| = \rho(A)$. Then $\lambda = \rho(A)$ and $\{u\}$ is a basis for $E_\lambda$, where $u \in \mathsf{C}^n$ is the column vector in which each coordinate equals 1.
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\C)$ be a matrix in which each entry is positive, and let $\lambda$ be an eigenvalue of $A$ such that $|\lambda| = \nu(A)$. Then $\lambda = \nu(A)$, and the dimension of $E_\lambda = 1$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Let $A \in M_{n \times n}(\C)$ be a transition matrix in which each entry is positive, and let $\lambda$ be any eigenvalue of $A$ other than $1$. Then $|\lambda| < 1$. Moreover, the eigenspace corresponding to the eigenvalue $1$ has dimension $1$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{theorem}
|
||||
\hfill\\
|
||||
Let $A$ be a regular transition matrix, and let $\lambda$ be an eigenvalue of $A$. Then
|
||||
|
||||
\begin{enumerate}
|
||||
\item $|\lambda| \leq 1$.
|
||||
\item If $|\lambda| = 1$, then $\lambda = 1$, and $\ldim{E_\lambda} = 1$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
\hfill\\
|
||||
Let $A$ be a regular transition matrix that is diagonalizable. Then $\displaystyle\lim_{m \to \infty}A^m$ exists.
|
||||
\end{corollary}
|
||||
|
||||
\begin{theorem}\label{Theorem 5.20}
|
||||
\hfill\\
|
||||
Let $A$ be an $n \times n$ regular transition matrix. Then
|
||||
|
||||
\begin{enumerate}
|
||||
\item The multiplicity of $1$ as an eigenvalue of $A$ is $1$.
|
||||
\item $\displaystyle\lim_{m \to \infty}A^m$ exists.
|
||||
\item $\L = \displaystyle\lim_{m \to \infty}A^m$ is a transition matrix.
|
||||
\item $AL = LA = L$.
|
||||
\item The columns of $L$ are identical. In fact, each column of $L$ is equal to the unique probability vector $v$ that is also an eigenvector of $A$ corresponding to the eigenvalue $1$.
|
||||
\item For any probability vector $w$, $\displaystyle\lim_{m \to \infty}(A^mw) = v$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
The vector $v$ in \autoref{Theorem 5.20}(5) is called the \textbf{fixed probability vector} or \textbf{stationary vector} of the regular transition matrix $A$.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
Consider transition matrices that can be represented in the following form:
|
||||
|
||||
\[\begin{pmatrix}
|
||||
I & B \\
|
||||
O & \C
|
||||
\end{pmatrix}\]
|
||||
|
||||
where $I$ is an identity matrix and $O$ is a zero matrix. (Such transition matrices are not regular since the lower left block remains $O$ in any power of the matrix.) The states corresponding to the identity submatrix are called \textbf{absorbing states} because such a state is never left once it is entered. A Markov chain is called an \textbf{absorbing Markov chain} if it is possible to go from an arbitrary state into an absorbing state in a finite number of stages.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}
|
||||
\hfill\\
|
||||
For $A \in M_{n \times n}(\C)$, define $e^A = \displaystyle\lim_{m \to \infty}B_m$, where
|
||||
|
||||
\[B_m = I + A + \frac{A^2}{2!} + \dots + \frac{A^m}{m!}\]
|
||||
|
||||
Thus $e^A$ is the sum of the infinite series
|
||||
|
||||
\[I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \dots,\]
|
||||
|
||||
and $B_m$ is the $m$th partial sum of this series. (Note the analogy with the power series
|
||||
|
||||
\[e^a = 1 + a + \frac{a^2}{2!}+\frac{a^3}{3!}+\dots,\]
|
||||
which is valid for all complex numbers $a$.)
|
||||
\end{definition}
|
||||
|
||||
Reference in New Issue
Block a user