week 13v1

Author: annadavismath

RTH-0010/main.tex

@@ -500,7 +500,7 @@ \subsection*{Orthogonal Decomposition of $\vec{x}$}
 
 \section*{Practice Problems}
 \begin{problem}\label{prob:rref_way}
-Retry Example~\ref{fourier} using Gaussian elimination.  Which method seems easier to you?
+Retry Example~\ref{fourier} using Gaussian elimination.  %Which method seems easier to you?
 \end{problem}
 
 \begin{problem}\label{prob:vec_eq_0}
@@ -510,27 +510,26 @@ \section*{Practice Problems}
 Show that $\vec{v}=\vec{0}$.
 \end{problem}
 
-\emph{Problems \ref{OrthoProj1.1}-\ref{OrthoProj1.3}}
-
+\begin{problem}\label{OrthoProj1.1}
 Let $\vec{x} = \begin{bmatrix}1\\ -2\\ 1\\ 6\end{bmatrix}$ in $\RR^4$, and let $W = \mbox{span}\left(\begin{bmatrix}2\\ 1\\ 3\\ -4\end{bmatrix}, \begin{bmatrix}1\\ 2\\ 0\\ 1\end{bmatrix}\right)$.
 
-\begin{problem}\label{OrthoProj1.1}
+\begin{question}
 Compute $\mbox{proj}_W(\vec{x})$.
 
 Answer:  $$\frac{1}{10}\begin{bmatrix}\answer{-9}\\\answer{3}\\\answer{-21}\\\answer{33}\end{bmatrix}$$
-\end{problem}
+\end{question}
 
-\begin{problem}\label{OrthoProj1.2}
+\begin{question}
 Show that $\left\{\begin{bmatrix}1\\ 0\\ 2\\ -3\end{bmatrix}, \begin{bmatrix}4\\ 7\\ 1\\ 2\end{bmatrix}\right\}$ is another orthogonal basis of $W$.
-\end{problem}
+\end{question}
 
-\begin{problem}\label{OrthoProj1.3}
-Use the basis in Problem \ref{OrthoProj1.2} to compute $\mbox{proj}_W(\vec{x})$.
+\begin{question}
+Use the basis you found in the previous part to compute $\mbox{proj}_W(\vec{x})$.
 
 Answer:  $$\frac{1}{70}\begin{bmatrix}\answer{-63}\\\answer{21}\\\answer{-147}\\\answer{231}\end{bmatrix}$$
+\end{question}
 \end{problem}
 
-
 \begin{problem}\label{prob:proofCor}
 Prove Corollary \ref{cor:orthProjOntoW}
 \end{problem}

RTH-0030/main.tex

@@ -92,7 +92,7 @@ \subsection*{Best Approximate Solutions}
 
 What we observed above, holds in general.  We will use this fact to find $\vec{z}$.
 
-Every vector in $\text{col}(A)$ can be written in the form $A\vec{x}$ for some $\vec{x}$ in $\RR^m$.  Our goal is to find $\vec{z}$ such that $A\vec{z}$ is the orthogonal projection of $\vec{b}$ onto $\text{col}(A)$. By Corollary \ref{cor:orthProjOntoW}, every vector $A\vec{x}$ in $\text{col}(A)$ is orthogonal to $\vec{b}-A\vec{z}$.  This means $\vec{b}-A\vec{z}$ is in the orthogonal complement of $\text{col}(A)$, which is $\text{null}(A^T)$.
+Every vector in $\text{col}(A)$ can be written in the form $A\vec{x}$ for some $\vec{x}$ in $\RR^m$.  Our goal is to find $\vec{z}$ such that $A\vec{z}$ is the orthogonal projection of $\vec{b}$ onto $\text{col}(A)$. By Corollary \ref{cor:orthProjOntoW}, every vector $A\vec{x}$ in $\text{col}(A)$ is orthogonal to $\vec{b}-A\vec{z}$.  %This means $\vec{b}-A\vec{z}$ is in the orthogonal complement of $\text{col}(A)$, which is $\text{null}(A^T)$.
 
 Therefore, we have
 %{\begin{eqnarray*}
@@ -418,11 +418,9 @@ \section*{Practice Problems}
 $$x=\answer{\frac{-20}{12}},\quad y=\answer{\frac{46}{12}},\quad z=\answer{\frac{95}{12}}$$
 \end{problem}
 
-\emph{Problems \ref{prob:leastSq2a}-\ref{prob:leastSq2b}}
-
-Find a linear function of best fit for each of the following sets of data points.  Examine how well your line fits the points by typing the equation of the line into the Desmos window.
 
 \begin{problem}\label{prob:leastSq2a}
+Find a linear function of best fit for the given set of data points.  Examine how well your line fits the points by typing the equation of the line into the Desmos window.
 $$(2,4), (4,3), (7,2), (8,1)$$
 Enter your answers in fraction form.
 $$f(x)=\answer{\frac{-6}{13}}x+\answer{\frac{64}{13}}$$
@@ -434,6 +432,7 @@ \section*{Practice Problems}
 \end{problem}
 
 \begin{problem}\label{prob:leastSq2b}
+Find a linear function of best fit for the given set of data points.  Examine how well your line fits the points by typing the equation of the line into the Desmos window.
 $$(-2, 3), (-1,1), (0,0), (1, -2), (2, -4)$$
 
 $$f(x)=\answer{-1.7}x+\answer{-0.4}$$
@@ -449,21 +448,21 @@ \section*{Practice Problems}
 $$(-2,1),(0,0),(3,2),(4,3)$$
 Round your answers to three decimal places.
 $$f(x)=\answer{0.194}x^2+\answer{-0.024}x+\answer{0.127}$$
-For more information about doing least squares with Octave, please see \href{https://ximera.osu.edu/linearalgebradzv3/xOctave/OCT_orth/main}{Octave for Chapter 9}.
+For more information about doing least squares with Octave, please see \href{https://ximera.osu.edu/corela/xOctave/OCT_orth/main}{Octave for Chapter 9}.
 \end{problem}
 
-\begin{problem}\label{ex:5_6_14}
-If $A$ is an $m \times n$ matrix, it can be proved that there exists a unique $n \times m$ matrix $A^{\#}$ satisfying the following four conditions: $AA^{\#}A = A$; $A^{\#}AA^{\#} = A^{\#}$; $AA^{\#}$ and $A^{\#}A$ are symmetric. The matrix $A^{\#}$ is called the \emph{Moore-Penrose} inverse.
+% \begin{problem}\label{ex:5_6_14}
+% If $A$ is an $m \times n$ matrix, it can be proved that there exists a unique $n \times m$ matrix $A^{\#}$ satisfying the following four conditions: $AA^{\#}A = A$; $A^{\#}AA^{\#} = A^{\#}$; $AA^{\#}$ and $A^{\#}A$ are symmetric. The matrix $A^{\#}$ is called the \emph{Moore-Penrose} inverse.
 
-\begin{enumerate}
-\item If $A$ is square and invertible, show that $A^{\#} = A^{-1}$.
+% \begin{enumerate}
+% \item If $A$ is square and invertible, show that $A^{\#} = A^{-1}$.
 
-\item If $\text{rank} A = m$, show that $A^{\#} = A^{T}(AA^{T})^{-1}$.
+% \item If $\text{rank} A = m$, show that $A^{\#} = A^{T}(AA^{T})^{-1}$.
 
-\item If $\text{rank} A = n$, show that $A^{\#} = (A^{T}A)^{-1}A^{T}$.  (Notice the appearance of the Moore-Penrose inverse arrived when we solve the normal equations, arriving at Equation (\ref{eq:leastSquaresZ})). 
+% \item If $\text{rank} A = n$, show that $A^{\#} = (A^{T}A)^{-1}A^{T}$.  (Notice the appearance of the Moore-Penrose inverse arrived when we solve the normal equations, arriving at Equation (\ref{eq:leastSquaresZ})). 
 
-\end{enumerate}
-\end{problem}
+% \end{enumerate}
+% \end{problem}
 
 
 % In many scientific investigations, data are collected that relate two variables. For example, if $x$ is the

RTH-0035/main.tex

@@ -21,7 +21,7 @@ \section*{Orthogonal Matrices and Symmetric Matrices}
 As we have seen, the nice bases of $\RR^n$ are the orthogonal ones, so a natural question is: which $n \times n$ matrices have $n$ orthogonal eigenvectors, so that columns of $P$ form an orthogonal basis for $\RR^n$? These turn out to be precisely the \dfn{symmetric matrices} (matrices for which $A=A^T$), and this is the main result of this section.
 
 \section*{Orthogonal Matrices}
-Recall that an orthogonal set of vectors is called \dfn{orthonormal} if $\norm{\vec{q}} = 1$ for each vector $\vec{q}$ in the set, and that any orthogonal set $\{\vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{k}\}$ can be ``\textit{normalized}'', i.e. converted into an orthonormal set $\left\{ \frac{1}{\norm{\vec{v}_{1}}}\vec{v}_{1}, \frac{1}{\norm{\vec{v}_{2}}}\vec{v}_{2}, \dots, \frac{1}{\norm{\vec{v}_{k}}}\vec{v}_{k} \right\}$. In particular, if a matrix $A$ has $n$ orthogonal eigenvectors, they can (by normalizing) be taken to be orthonormal. The corresponding diagonalizing matrix (we will use $Q$ instead of $P$) has orthonormal columns, and such matrices are very easy to invert.
+A collection of non-zero, pairwise orthogonal vectors in $\RR^n$ is called an \dfn{orthogonal} set of vectors. An orthogonal set of vectors is called \dfn{orthonormal} if $\norm{\vec{q}} = 1$ for each vector $\vec{q}$ in the set. A set of orthogonal vectors $\{\vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{k}\}$ can be ``\textit{normalized}'', i.e. converted into an orthonormal set $\left\{ \frac{1}{\norm{\vec{v}_{1}}}\vec{v}_{1}, \frac{1}{\norm{\vec{v}_{2}}}\vec{v}_{2}, \dots, \frac{1}{\norm{\vec{v}_{k}}}\vec{v}_{k} \right\}$. In particular, if a matrix $A$ has $n$ orthogonal eigenvectors, they can (by normalizing) be taken to be orthonormal. The corresponding diagonalizing matrix (we will use $Q$ instead of $P$) has orthonormal columns, and such matrices are very easy to invert.
 
 
 \begin{theorem}\label{th:orthogonal_matrices}
@@ -263,7 +263,7 @@ \section*{Symmetric Matrices}
 \end{proof}
 
 
-Because the eigenvalues of a real symmetric matrix are real, Theorem~\ref{th:PrinAxes} is also called the \dfn{Real Spectral Theorem}, and the set of distinct eigenvalues is called the \dfn{spectrum} of the matrix. A similar result holds for matrices with complex entries (Theorem \ref{th:025890}).
+Because the eigenvalues of a real symmetric matrix are real, Theorem~\ref{th:PrinAxes} is called the \dfn{Real Spectral Theorem}, and the set of distinct eigenvalues is called the \dfn{spectrum} of the matrix. A similar result holds for matrices with complex entries (Theorem \ref{th:025890}).
 
 
 \begin{example}\label{ex:DiagonalizeSymmetricMatrix}
@@ -274,13 +274,13 @@ \section*{Symmetric Matrices}
 \end{bmatrix}$.
 
 \begin{explanation}
- The characteristic polynomial of $A$ is (adding twice row 1 to row 2):
+ The characteristic equation of $A$ is (adding twice row 1 to row 2):
 \begin{equation*}
-c_{A}(z) = \det \begin{bmatrix}
-z - 1 & 0 & 1 \\
-0 & z - 1 & -2 \\
-1 & -2 & z - 5
-\end{bmatrix} = z(z - 1)(z - 6)
+\det \begin{bmatrix}
+1-\lambda & 0 & 1 \\
+0 & 1-\lambda & -2 \\
+1 & -2 & 5-\lambda
+\end{bmatrix} = \lambda(\lambda - 1)(\lambda - 6)=0
 \end{equation*}
 Thus the eigenvalues are $\lambda = 0$, $1$, and $6$, and corresponding eigenvectors are
 \begin{equation*}
@@ -371,13 +371,13 @@ \section*{Symmetric Matrices}
 
 
 \begin{explanation}
-  The characteristic polynomial is
+  The characteristic equation is
 \begin{equation*}
-c_{A}(z) = \det  \begin{bmatrix}
-z-8 & 2 & -2 \\
-2 & z-5 & -4 \\
--2 & -4 & z-5
-\end{bmatrix} = z(z-9)^2
+\det  \begin{bmatrix}
+8-\lambda & 2 & -2 \\
+2 & 5-\lambda & -4 \\
+-2 & -4 & 5-\lambda
+\end{bmatrix} = \lambda(\lambda-9)^2=0
 \end{equation*}
 Hence the distinct eigenvalues are $0$ and $9$ are of algebraic multiplicity $1$ and $2$, respectively.  The geometric multiplicities must be the same, for $A$ is diagonalizable, being symmetric. It follows that $\mbox{dim}(\mathcal{S}_0) = 1$ and $\mbox{dim}(\mathcal{S}_9) = 2$.  Gaussian elimination gives
 \begin{equation*}
@@ -395,7 +395,7 @@ \section*{Symmetric Matrices}
 1
 \end{bmatrix} \right\rbrace
 \end{equation*}
-The eigenvectors in $\mathcal{S}_{9}$ are both orthogonal to $\vec{x}_{1}$ as Theorem~\ref{th:symmetric_has_ortho_ev} guarantees, but not to each other. However, the Gram-Schmidt process yields an orthogonal basis
+The eigenvectors in $\mathcal{S}_{9}$ are both orthogonal to $\vec{x}_{1}$ as Theorem~\ref{th:symmetric_has_ortho_ev} guarantees, but not to each other. However, an orthogonal basis can be found using projections. (See \href{https://ximera.osu.edu/corela/LinearAlgebraInteractiveIntro/RTH-0015/main}{Gram-Schmidt Orthogonalization})
 \begin{equation*}
 \{\vec{f}_{2}, \vec{f}_{3}\} \mbox{ of } \mathcal{S}_{9}(A) \quad \mbox{ where } \quad \vec{f}_{2} = \begin{bmatrix}
 -2 \\
@@ -493,58 +493,66 @@ \section*{Symmetric Matrices}
 
 \section*{Practice Problems}
 
-\begin{problem}\label{prob:ortho_diag_implies_symmetric}
-Suppose $A$ is orthogonally diagonalizable.  Prove that $A$ is symmetric.  (This is the easy direction of the "if and only if" in Theorem \ref{th:PrinAxes}.)
-\end{problem}
-
-\emph{Problems \ref{prob:make_ortho_matrix3}-\ref{prob:make_ortho_matrix7}}
-
-Normalize the rows to make each of the following matrices orthogonal.
-
-
 \begin{problem}\label{prob:make_ortho_matrix3}
+Normalize the columns to make the following matrix orthogonal.
 $A = \begin{bmatrix}
-1 & 2 \\
--4 & 2
+1 & -4 \\
+2 & 2
 \end{bmatrix}$
+
+$$\frac{1}{\sqrt{\answer{5}}}\begin{bmatrix}\answer{1} & \answer{-2}\\\answer{2} & \answer{1}\end{bmatrix}$$
 \end{problem}
 
 \begin{problem}\label{prob:make_ortho_matrix7}
+Normalize the columns to make the following matrix orthogonal.
 $A = \begin{bmatrix}
 -1 & 2 & 2 \\
 2 & -1 & 2 \\
 2 & 2 & -1
 \end{bmatrix}$
+$$\frac{1}{\answer{3}}\begin{bmatrix}\answer{-1} & \answer{2} & \answer{2} \\
+\answer{2} & \answer{-1} & \answer{2} \\
+\answer{2} & \answer{2} & \answer{-1}
+\end{bmatrix}$$
 \end{problem}
 
 \begin{problem}\label{prob:findQ}
 For each matrix $A$, find an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.
 
-\begin{enumerate}
-\item\label{prob:findQa} $A = \begin{bmatrix}
+\begin{question}
+$$A = \begin{bmatrix}
 0 & 1 \\
 1 & 0
-\end{bmatrix}$
+\end{bmatrix}$$
+$$Q=\frac{1}{\sqrt{\answer{2}}}\begin{bmatrix}1 & -1\\\answer{1} & \answer{1}\end{bmatrix}$$
+\end{question}
 
-\item\label{prob:findQd} $A = \begin{bmatrix}
+\begin{question}
+    $$A = \begin{bmatrix}
 3 & 0 & 7 \\
 0 & 5 & 0 \\
 7 & 0 & 3
-\end{bmatrix}$
-
-\item\label{prob:findQe} $A = \begin{bmatrix}
-1 & 1 & 0 \\
-1 & 1 & 0 \\
-0 & 0 & 2
-\end{bmatrix}$
+\end{bmatrix}$$
+List eigenvalues in increasing order.
+$$\lambda_1=\answer{-4},\quad\lambda_2=\answer{5},\quad\lambda_3=\answer{10}$$
+$$Q^{-1}AQ=\begin{bmatrix}\answer{-1}/\sqrt{\answer{2}} & \answer{0} & \answer{1}/\sqrt{\answer{2}}\\\answer{0} & \answer{1} & \answer{0}\\1/\sqrt{\answer{2}} & \answer{0} & 1/\sqrt{\answer{2}}\end{bmatrix}\begin{bmatrix}\lambda_1 & 0 & 0\\0 & \lambda_2 & 0\\0 & 0& \lambda_3\end{bmatrix}\begin{bmatrix}\answer{-1}/\sqrt{\answer{2}} & \answer{0} & \answer{1}/\sqrt{\answer{2}}\\\answer{0} & \answer{1} & \answer{0}\\1/\sqrt{\answer{2}} & \answer{0} & 1/\sqrt{\answer{2}}\end{bmatrix}$$
+\end{question}
+
+% \begin{question}
+% $$A = \begin{bmatrix}
+% 1 & 1 & 0 \\
+% 1 & 1 & 0 \\
+% 0 & 0 & 2
+% \end{bmatrix}$$
+% \end{question}
+
+% (challenging problem) $A = \begin{bmatrix}
+% 3 & 5 & -1 & 1 \\
+% 5 & 3 & 1 & -1 \\
+% -1 & 1 & 3 & 5 \\
+% 1 & -1 & 5 & 3
+% \end{bmatrix}$
 
-\item\label{prob:findQg} (challenging problem) $A = \begin{bmatrix}
-3 & 5 & -1 & 1 \\
-5 & 3 & 1 & -1 \\
--1 & 1 & 3 & 5 \\
-1 & -1 & 5 & 3
-\end{bmatrix}$
-\end{enumerate}
 
 %\begin{sol}
 %\begin{enumerate} 
@@ -584,6 +592,9 @@ \section*{Practice Problems}
 %\end{sol}
 \end{problem}
 
+\begin{problem}\label{prob:ortho_diag_implies_symmetric}
+Suppose $A$ is orthogonally diagonalizable.  Prove that $A$ is symmetric.  (This is the easy direction of the "if and only if" in Theorem \ref{th:PrinAxes}.)
+\end{problem}
 
 \begin{problem}\label{ex:8_2_15}
 Prove the converse of Theorem~\ref{th:dotpSymmetric}:
@@ -605,16 +616,16 @@ \section*{Practice Problems}
 
 \item Give $2 \times 2$ examples of $Q$ such that $\det Q = 1$ and $\det  Q = -1$.
 
-\item If $\det  Q = -1$, show that $I + Q$ has no inverse.
-\begin{hint}
-$Q^{T}(I + Q) = (I + Q)^{T}$.
-\end{hint}
+% \item If $\det  Q = -1$, show that $I + Q$ has no inverse.
+% \begin{hint}
+% $Q^{T}(I + Q) = (I + Q)^{T}$.
+% \end{hint}
 
-\item If $P$ is $n \times n$ and $\det P \neq (-1)^{n}$, show that $I - P$ has no inverse.
+% \item If $P$ is $n \times n$ and $\det P \neq (-1)^{n}$, show that $I - P$ has no inverse.
 
-\begin{hint}
-$P^{T}(I - P) = -(I - P)^{T}$
-\end{hint}
+% \begin{hint}
+% $P^{T}(I - P) = -(I - P)^{T}$
+% \end{hint}
 \end{enumerate}
 %\begin{sol}
 %\begin{enumerate} 
@@ -638,41 +649,41 @@ \section*{Practice Problems}
 \end{problem}
 
 
-\begin{problem}\label{prob:ortho21}
-A matrix that we obtain from the identity matrix by writing its rows in a different order is called a \dfn{permutation matrix} (see Theorem \ref{th:LUPA}). Show that every permutation matrix is orthogonal.
-\end{problem}
+% \begin{problem}\label{prob:ortho21}
+% A matrix that we obtain from the identity matrix by writing its rows in a different order is called a \dfn{permutation matrix} (see Theorem \ref{th:LUPA}). Show that every permutation matrix is orthogonal.
+% \end{problem}
 
 
-\begin{problem}\label{prob:ortho25}
-Show that the following are equivalent for an $n \times n$ matrix $Q$.
+% \begin{problem}\label{prob:ortho25}
+% Show that the following are equivalent for an $n \times n$ matrix $Q$.
 
 
-\begin{enumerate} 
-\item $Q$ is orthogonal.
+% \begin{enumerate} 
+% \item $Q$ is orthogonal.
 
-\item $\norm{Q\vec{x}} = \norm{\vec{x}}$ for all $\vec{x}\in\RR^n$.
+% \item $\norm{Q\vec{x}} = \norm{\vec{x}}$ for all $\vec{x}\in\RR^n$.
 
-\item $\norm{ Q\vec{x} - Q\vec{y}} = \norm{\vec{x} - \vec{y}}$ for all $\vec{x}$, $\vec{y}\in \RR^n$.
+% \item $\norm{ Q\vec{x} - Q\vec{y}} = \norm{\vec{x} - \vec{y}}$ for all $\vec{x}$, $\vec{y}\in \RR^n$.
 
-\item $(Q\vec{x}) \dotp (Q\vec{y}) = \vec{x} \dotp \vec{y}$ for all columns $\vec{x}$, $\vec{y}\in\RR^n$.
+% \item $(Q\vec{x}) \dotp (Q\vec{y}) = \vec{x} \dotp \vec{y}$ for all columns $\vec{x}$, $\vec{y}\in\RR^n$.
 
-\begin{hint}
-%For (c) $\Rightarrow$ (d), see Exercise \ref{ex:5_3_14}(a). 
-For (d) $\Rightarrow$ (a), show that column $i$ of $Q$ equals $Q\vec{e}_{i}$, where $\vec{e}_{i}$ is column $i$ of the identity matrix.
-\end{hint}
-\end{enumerate}
+% \begin{hint}
+% %For (c) $\Rightarrow$ (d), see Exercise \ref{ex:5_3_14}(a). 
+% For (d) $\Rightarrow$ (a), show that column $i$ of $Q$ equals $Q\vec{e}_{i}$, where $\vec{e}_{i}$ is column $i$ of the identity matrix.
+% \end{hint}
+% \end{enumerate}
 
-\begin{remark}
-    This exercise shows that linear transformations with orthogonal standard matrices are distance-preserving (b,c) and angle-preserving (d).
-\end{remark}
+% \begin{remark}
+%     This exercise shows that linear transformations with orthogonal standard matrices are distance-preserving (b,c) and angle-preserving (d).
+% \end{remark}
 
-\end{problem}
+% \end{problem}
 
 
 
 
-\begin{problem}\label{prob:SchurChallenge}
-Modify the proof of Theorem~\ref{th:PrinAxes} to prove Theorem \ref{th:Schur}.
+% \begin{problem}\label{prob:SchurChallenge}
+% Modify the proof of Theorem~\ref{th:PrinAxes} to prove Theorem \ref{th:Schur}.
 %If $A\vec{x}_{1} = \lambda_{1}\vec{x}_{1}$ where $\norm{\vec{x}_{1}} = 1$, let $\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{n}\}$ be an orthonormal basis of $\RR^n$, and let $P_{1} = \begin{bmatrix}
 %\vec{x}_{1} & \vec{x}_{2} & \cdots &  \vec{x}_{n}
 %\end{bmatrix}$. Then $P_{1}$ is orthogonal and $P_{1}^TAP_{1} = \begin{bmatrix}
@@ -686,7 +697,7 @@ \section*{Practice Problems}
 %0 & T_{1}
 %\end{bmatrix}$
 % is upper triangular.
-\end{problem}
+% \end{problem}
 
 \section*{Text Source} This section was adapted from Section 8.2 of Keith Nicholson's \href{https://open.umn.edu/opentextbooks/textbooks/linear-algebra-with-applications}{\it Linear Algebra with Applications}. (CC-BY-NC-SA)