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.

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2024-02-21 19:31:36 -07:00
parent 85f421fca3
commit 09c5be15d7
45 changed files with 2602 additions and 379 deletions
@@ -3,13 +3,13 @@
\begin{definition}
\hfill\\
Let $A$ be an $m \times n$ matrix. Any one of the following three operations on the rows [columns] of $A$ is called an \textbf{elementary row [column] operation}:
\begin{enumerate}
\item interchanging any two rows [columns] of $A$;
\item multiplying any row [column] of $A$ by a nonzero scalar;
\item adding any scalar multiple of a row [column] of $A$ to another row [column].
\end{enumerate}
Any of these three operations are called an \textbf{elementary operation}. Elementary operations are of \textbf{type 1}, \textbf{type 2}, or \textbf{type 3} depending on whether they are obtained by (1), (2), or (3).
\end{definition}
@@ -26,4 +26,4 @@
\begin{theorem}
\hfill\\
Elementary matrices are invertible, and the inverse of an elementary matrix is an elementary matrix of the same type.
\end{theorem}
\end{theorem}
@@ -18,12 +18,12 @@
\begin{definition}
\hfill\\
A matrix is said to be in \textbf{reduced row echelon form} if the following three conditions are satisfied.
\begin{enumerate}
\item Any row containing a nonzero entry precedes any row in which all the entries are zero (if any).
\item The first nonzero entry in each row is the only nonzero entry in its column.
\item The first nonzero entry in each row is 1 and it occurs in a column to the right of the first nonzero entry in the preceding row.
\end{enumerate}
\end{definition}
@@ -31,10 +31,10 @@
\begin{definition}
\hfill\\
The following procedure for reducing an augmented matrix to reduced row echelon form is called \textbf{Gaussian elimination}. It consists of two separate parts.
\begin{enumerate}
\item In the \textit{forward pass}, the augmented matrix is transformed into an upper triangular matrix in which the first nonzero entry of each row is $1$, and it occurs in a column to the right of the first nonzero entry in the preceding row.
\item In the \textit{backward pass} or \textit{back-substitution}, the upper triangular matrix is transformed into reduced row echelon form by making the first nonzero entry of each row the only nonzero entry of its column.
\end{enumerate}
\end{definition}
@@ -46,30 +46,30 @@
\begin{definition}
A solution to a system of equations of the form
\[s = s_0 + t_1u_1 + t_2u_2 + \dots +t_{n-r}u_{n-r},\]
where $r$ is the number of nonzero solutions in $A'$ ($r \leq m$), is called a \textbf{general solution} of the system $Ax = b$. It expresses an arbitrary solution $s$ of $Ax = b$ in terms of $n - r$ parameters.
\end{definition}
\begin{theorem}
\hfill\\
Let $Ax = b$ be a system of $r$ nonzero equations in $n$ unknowns. Suppose that $\rank{A} = \rank{A|b}$ and that $(A|b)$ is in reduced row echelon form. Then
\begin{enumerate}
\item $\rank{A} = r$.
\item If the general solution obtained by the procedure above is of the form
\[s = s_0 + t_1u_1 + t_2u_2 + \dots + t_{n-r}u_{n-r},\]
then $\{u_1, u_2, \dots, u_{n-r}\}$ is a basis for the solution set of the corresponding homogeneous system, and $s_0$ is a solution to the original system.
\[s = s_0 + t_1u_1 + t_2u_2 + \dots + t_{n-r}u_{n-r},\]
then $\{u_1, u_2, \dots, u_{n-r}\}$ is a basis for the solution set of the corresponding homogeneous system, and $s_0$ is a solution to the original system.
\end{enumerate}
\end{theorem}
\begin{theorem}
\hfill\\
Let $A$ be an $m \times n$ matrix of rank $r$, where $r > 0$, and let $B$ be the reduced row echelon form of $A$. Then
\begin{enumerate}
\item The number of nonzero rows in $B$ is $r$.
\item For each $i = 1, 2, \dots, r$, there is a column $b_{j_i}$ of $B$ such that $b_{j_i} = e_i$.
@@ -81,4 +81,4 @@
\begin{corollary}
\hfill\\
The reduced row echelon form of a matrix is unique.
\end{corollary}
\end{corollary}
@@ -2,8 +2,8 @@
\begin{definition}
\hfill\\
The system of equations
The system of equations
\begin{equation}\label{eq:S}
\tag{S}
\begin{split}
@@ -13,47 +13,47 @@
a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m,
\end{split}
\end{equation}
where $a_{ij}$ and $b_i$ ($1 \leq i \leq m$ and $1 \leq j \leq n$) are scalars in a field $\F$ and $x_1, x_2, \dots, x_n$ are $n$ variables taking values in $\F$, is a called a \textbf{system of $m$ linear equations in $n$ unknowns over the field $\F$}.
The $m \times n$ matrix
The $m \times n$ matrix
\[\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{m1} & a_{m2} & \dots & a_{mn}
\end{pmatrix}\]
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{m1} & a_{m2} & \dots & a_{mn}
\end{pmatrix}\]
is called the \textbf{coefficient matrix} of the system \eqref{eq:S}.
If we let
\[x = \begin{pmatrix}
x_1 \\ x_2 \\ \vdots \\ x_n
\end{pmatrix}\ \ \text{and}\ \ b = \begin{pmatrix}
b_1 \\ b_2 \\ \vdots \\ b_m
\end{pmatrix},\]
x_1 \\ x_2 \\ \vdots \\ x_n
\end{pmatrix}\ \ \text{and}\ \ b = \begin{pmatrix}
b_1 \\ b_2 \\ \vdots \\ b_m
\end{pmatrix},\]
then the system \eqref{eq:S} may be rewritten as a single matrix equation
\[Ax = b.\]
To exploit the results that we have developed, we often consider a system of linear equations as a single matrix equation.
A \textbf{solution} to the system \eqref{eq:S} is an $n$-tuple
\[s = \begin{pmatrix}
s_1 \\ s_2 \\ \vdots \\ s_n
\end{pmatrix} \in \F^n\]
s_1 \\ s_2 \\ \vdots \\ s_n
\end{pmatrix} \in \F^n\]
such that $As = b$. The set of all solutions to the system \eqref{eq:S} is called the \textbf{solution set} of the system. System \eqref{eq:S} is called \textbf{consistent} if its solution set is nonempty; otherwise it is called \textbf{inconsistent}.
\end{definition}
\begin{definition}
\hfill\\
A system $Ax = b$ of $m$ linear equations in $n$ unknowns is said to be \textbf{homogeneous} if $b = 0$. Otherwise the system is said to be \textbf{nonhomogeneous}.\\
Any homogeneous system has at least one solution, namely, the zero vector.
\end{definition}
@@ -75,7 +75,7 @@
\begin{theorem}
\hfill\\
Let $K$ be the solution set of a system of linear equations $Ax = b$, and let $\mathsf{K}_\mathsf{H}$ be the solution set of the corresponding homogeneous system $Ax = 0$. Then for any solution $s$ to $Ax = b$
\[K = \{s\} + \mathsf{K}_\mathsf{H} = \{s + k: k \in \mathsf{K}_\mathsf{H}\}.\]
\end{theorem}
@@ -96,33 +96,33 @@
\begin{definition}
Consider a system of linear equations
\[\begin{split}
a_{11}p_1 + a_{12}p_2 + \dots + a_{1m}p_m = p_1 \\
a_{21}p_1 + a_{22}p_2 + \dots + a_{2m}p_m = p_2 \\
\dots \\
a_{n1}p_1 + a_{n2}p_2 + \dots + a_{nm}p_m = p_m \\
\end{split}\]
This system can be written as $Ap = p$, where
a_{11}p_1 + a_{12}p_2 + \dots + a_{1m}p_m = p_1 \\
a_{21}p_1 + a_{22}p_2 + \dots + a_{2m}p_m = p_2 \\
\dots \\
a_{n1}p_1 + a_{n2}p_2 + \dots + a_{nm}p_m = p_m \\
\end{split}\]
This system can be written as $Ap = p$, where
\[p = \begin{pmatrix}
p_1 \\ p_2 \\ \vdots \\ p_m
\end{pmatrix}\]
and $A$ is the coefficient matrix of the system. In this context, $A$ is called the \textbf{input-ouput (or consumption) matrix}, and $Ap = p$ is called the \textbf{equilibrium condition}.
For vectors $b = (b_1, b_2, \dots, b_n)$ and $c = (c_1, c_2, \dots, c_n)$ in $\R^n$, we use the notation $b \geq c$ [$b > c$] to mean $b_i \geq c_i$ [$b_i > c_i$] for all $i$. The vector $b$ is called \textbf{nonnegative [positive]} if $b \geq 0$ [$b > 0$].
p_1 \\ p_2 \\ \vdots \\ p_m
\end{pmatrix}\]
and $A$ is the coefficient matrix of the system. In this context, $A$ is called the \textbf{input-output (or consumption) matrix}, and $Ap = p$ is called the \textbf{equilibrium condition}.
For vectors $b = (b_1, b_2, \dots, b_n)$ and $c = (c_1, c_2, \dots, c_n)$ in $\R^n$, we use the notation $b \geq c$ [$b > c$] to mean $b_i \geq c_i$ [$b_i > c_i$] for all $i$. The vector $b$ is called \textbf{non-negative [positive]} if $b \geq 0$ [$b > 0$].
\end{definition}
\begin{theorem}
\hfill\\
Let $A$ be an $n \times n$ input-output matrix having the form
\[A = \begin{pmatrix}
B & C \\
D & E
\end{pmatrix},\]
where $D$ is a $1 \times (n -1)$ positive vector and $C$ is an $(n-1)\times 1$ positive vector. Then $(I -A)x = 0$ has a one-dimensional solution set that is generated by a nonnegative vector.
\end{theorem}
B & C \\
D & E
\end{pmatrix},\]
where $D$ is a $1 \times (n -1)$ positive vector and $C$ is an $(n-1)\times 1$ positive vector. Then $(I -A)x = 0$ has a one-dimensional solution set that is generated by a non-negative vector.
\end{theorem}
@@ -13,7 +13,7 @@
\begin{theorem}
\hfill\\
Let $A$ be an $m \times n$ matrix. if $P$ and $Q$ are invertible $m \times m$ and $n \times n$ matrices, respectively, then
\begin{enumerate}
\item $\rank{AQ} = \rank{A}$,
\item $\rank{PA} = \rank{A}$,\\ and therefore
@@ -34,30 +34,30 @@
\begin{theorem}
\hfill\\
Let $A$ be an $m \times n$ matrix of rank $r$. Then $r \leq m$, $r \leq n$, and, by means of a finite number of elementary row and column operations, $A$ can be transformed into the matrix
\[D = \begin{pmatrix}
I_r & O_1 \\
O_2 & O_3
\end{pmatrix}\]
I_r & O_1 \\
O_2 & O_3
\end{pmatrix}\]
where $O_1$, $O_2$ and $O_3$ are the zero matrices. Thus $D_{ii} = 1$ for $i \leq r$ and $D_{ij} = 0$ otherwise.
\end{theorem}
\begin{corollary}
\hfill\\
Let $A$ be an $m \times n$ matrix of rank $r$. Then there exist invertible matrices $B$ and $C$ of sizes $m \times m$ and $n \times n$, respectively, such that $D=BAC$, where
\[D = \begin{pmatrix}
I_r & O_1 \\
O_2 & O_3
\end{pmatrix}\]
I_r & O_1 \\
O_2 & O_3
\end{pmatrix}\]
is the $m \times n$ matrix in which $O_1$, $O_2$, and $O_3$ are zero matrices.
\end{corollary}
\begin{corollary}
\hfill\\
Let $A$ be an $m \times n$ matrix. Then
\begin{enumerate}
\item $\rank{A^t} = \rank{A}$.
\item The rank of any matrix equals the maximum number of its linearly independent rows; that is, the rank of a matrix is the dimension of the subspace generated by its rows.
@@ -73,7 +73,7 @@
\begin{theorem}
\hfill\\
Let $T: V \to W$ and $U: W \to Z$ be linear transformations on finite-dimensional vector spaces $V$, $W$, and $Z$, and let $A$ and $B$ be matrices such that the product $AB$ is defined. Then
\begin{enumerate}
\item $\rank{UT} \leq \rank{U}$.
\item $\rank{UT} \leq \rank{T}$.
@@ -85,4 +85,4 @@
\begin{definition}
\hfill\\
Let $A$ and $B$ be $m \times n$ and $m \times p$ matrices, respectively. By the \textbf{augmented matrix} $(A|B)$, we mean the $m \times (n \times p)$ matrix $(A\ B)$, that is, the matrix whose first $n$ columns are the columns of $A$, and whose last $p$ columns are the columns of $B$.
\end{definition}
\end{definition}