week 12

Author: annadavismath

EIG-0020/main.tex

@@ -14,7 +14,7 @@
 \section*{The Characteristic Equation}
 \end{onlineOnly}
 
-Let $A$ be an $n \times n$ matrix.  In \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/EIG-0010/main}{Describing Eigenvalues and Eigenvectors Algebraically and Geometrically} we learned that the eigenvectors and eigenvalues of $A$ are vectors $\vec{x}$ and scalars $\lambda$ that satisfy the equation  
+Let $A$ be an $n \times n$ matrix.  In the previous section we learned that the eigenvectors and eigenvalues of $A$ are vectors $\vec{x}$ and scalars $\lambda$ that satisfy the equation  
 \begin{align}\label{def:eigen} A \vec{x} = \lambda \vec{x}\end{align}
 We listed a few reasons why we are interested in finding eigenvalues and eigenvectors, but we did not give any process for finding them.  In this section we will focus on a process which can be used for small matrices.  For larger matrices, the best methods we have are iterative methods, and we will explore some of these in \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/EIG-0070/main}{The Power Method and the Dominant Eigenvalue}.
 
@@ -29,7 +29,7 @@ \section*{The Characteristic Equation}
 A\vec{x}-\lambda I\vec{x} &= \vec{0} \\
 (A-\lambda I)\vec{x} &= \vec{0}
 \end{align*}
-The middle step was necessary before factoring because \wordChoice{\choice[correct]{we cannot subtract a $1 \times 1$ scalar $\lambda$ from an $n \times n$ matrix $A$} \choice{$\lambda$ is a Greek letter}}.
+%The middle step was necessary before factoring because we cannot subtract a $1 \times 1$ scalar $\lambda$ from an $n \times n$ matrix $A$.
 
 This shows that any eigenvector $\vec{x}$ of $A$ is in the \wordChoice{\choice{row space}\choice{column space}\choice[correct]{null space}}  of the related matrix, $A-\lambda I$.
 
@@ -79,12 +79,18 @@ \subsection*{Eigenvalues}
 &=-(\lambda-4)(\lambda-1)^2
 \end{align*}
 Matrix $C$ has eigenvalues $\lambda_1=1$ and $\lambda_2=4$.
+
+\begin{remark}
+    The factor  $(\lambda-1)$ appears twice.  This repeated factor gives rise to the eigenvalue $\lambda_1=1$.  We say that the eigenvalue $\lambda_1=1$ has \dfn{algebraic multiplicity} $2$.
+\end{remark}
 \end{explanation}
 \end{example}
 
-In Example \ref{ex:3x3eig}, the factor  $(\lambda-1)$ appears twice.  This repeated factor gives rise to the eigenvalue $\lambda_1=1$.  We say that the eigenvalue $\lambda_1=1$ has \dfn{algebraic multiplicity} $2$. 
+\begin{definition}\label{def:algMult}
+    \dfn{Algebraic multiplicity} of an eigenvalue is its multiplicity as a root of the characteristic equation.
+\end{definition} 
 
-The three examples above are a bit contrived.  It is not always possible to completely factor the characteristic polynomial using only real numbers.  However, a fundamental fact from algebra is that every degree $n$ polynomial has $n$ roots (counting multiplicity) provided that we allow complex numbers.  This is why sometimes eigenvalues and their corresponding eigenvectors involve complex numbers.  The next example illustrates this point.
+It is not always possible to completely factor the characteristic polynomial using only real numbers.  However, a fundamental fact from algebra is that every degree $n$ polynomial has $n$ roots (counting multiplicity) provided that we allow complex numbers.  This is why sometimes eigenvalues and their corresponding eigenvectors involve complex numbers.  The next example illustrates this point.
 
 \begin{example}\label{ex:3x3_complex_eig}
 Let $D=\begin{bmatrix} 0&0&0\\ 0 &1&1\\ 0 & -1&1\end{bmatrix}$.  Compute the eigenvalues of this matrix.
@@ -137,10 +143,10 @@ \subsection*{Eigenvectors}
 \begin{align}\label{eqn:nullspace}
  (A-\lambda I) \vec{x}=\vec{0}   
 \end{align} 
-For any given eigenvalue $\lambda$ there are infinitely many eigenvectors associated with it.  In fact, the eigenvectors associated with $\lambda$ form a subspace of $\RR^n$. 
+As we have seen in the previous section, a scalar multiple of an eigenvector is an eigenvector.  So for any given eigenvalue $\lambda$ there are infinitely many eigenvectors associated with it.  In fact, the eigenvectors associated with $\lambda$, together with the zero vector, form a subspace of $\RR^n$. 
 
 \begin{theorem}\label{th:eigenspace}
-    Let $A$ be an $n\times n$ matrix and let $\lambda$ be an eigenvalue of $A$.  Then the set of all eigenvectors associated with $\lambda$ is a subspace of $\RR^n$.
+    Let $A$ be an $n\times n$ matrix and let $\lambda$ be an eigenvalue of $A$.  Then the set of all eigenvectors associated with $\lambda$, together with the zero vector, is a subspace of $\RR^n$.
 \end{theorem}
 \begin{proof}
     See Practice Problems \ref{prob:eigenspace1} and \ref{prob:eigenspace2}.
@@ -149,10 +155,10 @@ \subsection*{Eigenvectors}
 This motivates the following definition.
 
 \begin{definition}\label{def:eigspace}
-The set of all eigenvectors associated with a given eigenvalue of a matrix is known as the \dfn{eigenspace} associated with that eigenvalue.
+The set of all eigenvectors associated with a given eigenvalue of a matrix, together with the zero vector, is known as the \dfn{eigenspace} associated with that eigenvalue.
 \end{definition}
 
-So given an eigenvalue $\lambda$, there is an associated eigenspace $\mathcal{S}$, and our goal is to find a basis of $\mathcal{S}$, for then any eigenvector $\vec{x}$ will be a linear combination of the vectors in that basis.  Moreover, we are trying to find a basis for the set of vectors that satisfy Equation \ref{eqn:nullspace}, which means we seek a basis for $\mbox{null}(A-\lambda I)$.  We have already learned how to compute a basis of a null space - see \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/VSP-0040/main}{Subspaces Associated with Matrices}.
+Given an eigenvalue $\lambda$, there is an associated eigenspace $\mathcal{S}_{\lambda}$. Our goal is to find a basis of $\mathcal{S}_{\lambda}$.  Then any eigenvector $\vec{x}$ associated with $\lambda$ will be a linear combination of the vectors in that basis.  In seeking a basis for $\mathcal{S}_{\lambda}$, we are trying to find a basis for the set of vectors that satisfy $(A-\lambda I) \vec{x}=\vec{0}$, which means we seek a basis for $\mbox{null}(A-\lambda I)$. % We have already learned how to compute a basis of a null space - see \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/VSP-0040/main}{Subspaces Associated with Matrices}.
 
 Let's return to the examples we did in the first part of this section.
 
@@ -172,6 +178,14 @@ \subsection*{Eigenvectors}
 \begin{align*}\mbox{rref}(A-3I)=\mbox{rref}\left(\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\right)&=\begin{bmatrix}1&-1\\0&0\end{bmatrix},
 \end{align*}
 Vectors in the null space have the form $\begin{bmatrix}1\\1\end{bmatrix}t$ This means that $\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$ is one possible basis for the eigenspace $\mathcal{S}_3$.
+
+These computations reinforce the geometric insights into this problem that we gained in the previous section.
+
+\begin{onlineOnly}
+\begin{center}
+    \geogebra{yr2btbqj}{800}{600} 
+  \end{center}
+  \end{onlineOnly}
 \end{explanation}
 \end{example}
 
@@ -185,14 +199,10 @@ \subsection*{Eigenvectors}
 From this we see that an eigenvector in $\mathcal{S}_0$ has the form $\begin{bmatrix}-1/2\\1\end{bmatrix}t$. %$\begin{bmatrix}x_1\\x_2\end{bmatrix}$ in $\mathcal{S}_0$ satisfies $x_1+\frac{1}{2} x_2=0$, so that $2x_1=-x_2$. 
 This means that $\left\{\begin{bmatrix}-1/2\\1\end{bmatrix}\right\}$ is one possible basis for the eigenspace $\mathcal{S}_0$.  By letting $t=-2$, we obtain an arguably nicer-looking basis: $\left\{\begin{bmatrix}1\\-2\end{bmatrix}\right\}$. 
 
-See if you can compute a basis for $\mathcal{S}_4$.  
 
-Click on the arrow if you need help.
-
-\begin{expandable}{}{}
 To compute a basis for $\mathcal{S}_4$, the subspace of all eigenvectors associated to the eigenvalue $\lambda_2=4$, we compute:
 $$\mbox{rref}(B-4I)=\mbox{rref}\left(\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\right)=\begin{bmatrix}1&-\frac{1}{2}\\0&0\end{bmatrix}$$
-\end{expandable}
+
 
 From this we find that $\left\{\begin{bmatrix}1\\\answer{2}\end{bmatrix}\right\}$ is one possible basis for the eigenspace $\mathcal{S}_4$.
 \end{explanation}
@@ -216,9 +226,18 @@ \subsection*{Eigenvectors}
 \end{align*}
 This time there is one free variable.  %Setting $x_3=t$, we also get $x_1=t$ and $x_2=t$.  From this we see
 The eigenvectors in $\mathcal{S}_4$ have the form $\begin{bmatrix}t\\t\\t\end{bmatrix}$, so a possible basis for the eigenspace $\mathcal{S}_4$ is given by $\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}$.
+
+\begin{remark}
+    Recall that $\lambda_1=1$ has algebraic multiplicity $2$.  Note that $\text{dim}(\mathcal{S}_{\lambda_1})=2$.  We say that  $\lambda_1$ has geometric multiplicity $2$.  We have to be careful, however, not to assume that geometric and algebraic multiplicity are always the same.  We will revisit this issue later in the text.
+\end{remark}
+
 \end{explanation}
 \end{example}
 
+\begin{definition}\label{def:geomMult}
+    The \dfn{geometric multiplicity} of an eigenvalue is the dimension of the eigenspace corresponding to it.
+\end{definition}
+
 \begin{example}\label{ex:3x3_complex_ev} (Finding eigenvectors for Example \ref{ex:3x3_complex_eig})
 We know from Example \ref{ex:3x3_complex_eig} that $D=\begin{bmatrix} 0&0&0\\ 0 &1&1\\ 0 & -1&1\end{bmatrix}$ has eigenvalues $\lambda=0$, $\lambda_1=1+i$, and $\lambda_2=1-i$.  Compute a basis for the eigenspace associated with each eigenvalue.
 \begin{explanation}
@@ -243,39 +262,22 @@ \subsection*{Eigenvectors}
 \end{theorem}
 
 \begin{proof}
-A square matrix $A$ is singular if and only if $\det{A}=0$.(see  \ref{th:detofsingularmatrix}).  But $\det{A}=0$ if and only if $\det{A-0I}=0$, which is true if and only if zero is an eigenvalue of $A$.
+A square matrix $A$ is singular if and only if $\det{A}=0$. But $\det{A}=0$ if and only if $\det{A-0I}=0$, which is true if and only if zero is an eigenvalue of $A$.
 \end{proof}
 
 \section*{Practice Problems}
 
-\emph{Problems \ref{prob:eigenspace1}-\ref{prob:eigenspace2}}
-
-In this exercise we will prove that the eigenvectors associated with an eigenvalue $\lambda$ of an $n \times n$ matrix $A$ form a subspace of $\RR^n$.
-
-\begin{problem}\label{prob:eigenspace1}
-Let $\vec{x}$ and $\vec{y}$ be eigenvectors of $A$ associated with $\lambda$.  Show that $\vec{x}+\vec{y}$ is also an eigenvector of $A$ associated with $\lambda$.  (This shows that the set of eigenvectors of $A$ associated with $\lambda$ is closed under addition).
-\end{problem}
-
-\begin{problem}\label{prob:eigenspace2}
-Show that the set of eigenvectors of $A$ associated with $\lambda$ is closed under scalar multiplication.
-\end{problem}
-
-
-\emph{Problems \ref{prob:eigenspace3}-\ref{prob:eigenspace4}}
-
-Compute the eigenvalues of the given matrix and find the corresponding eigenspaces.
-
-\begin{problem}\label{prob:eigenspace3}
-$$\begin{bmatrix}4&1\\8&-3\end{bmatrix}$$
+\begin{problem}\label{prob:eigenspace3} Compute the eigenvalues of $A$, and find the corresponding eigenspaces.
+$$A=\begin{bmatrix}4&1\\8&-3\end{bmatrix}$$
 Answer:
 (List the eigenvalues in an increasing order.)
 $$\lambda_1=\answer{-4},\quad\lambda_2=\answer{5}$$
 
 A basis for $\mathcal{S}_{\lambda_1}$ is $\left\{\begin{bmatrix}\answer{-1/8}\\1\end{bmatrix}\right\}$.  A basis for $\mathcal{S}_{\lambda_2}$ is $\left\{\begin{bmatrix}\answer{1}\\1\end{bmatrix}\right\}$.
 \end{problem}
 
-\begin{problem}\label{prob:eigenspace4}
-$$\begin{bmatrix}1&-2\\2&1\end{bmatrix}$$
+\begin{problem}\label{prob:eigenspace4}Compute the eigenvalues of $A$, and find the corresponding eigenspaces.
+$$A=\begin{bmatrix}1&-2\\2&1\end{bmatrix}$$
 Answer:
 
 $$\lambda_1=\answer{1}+\answer{2}i,\quad\lambda_2=\answer{1}-\answer{2}i$$
@@ -288,12 +290,10 @@ \section*{Practice Problems}
 Let $T=\begin{bmatrix} 1 & 2 & 3\\ 0 & 5 & 6\\ 0 & 0 & 9\end{bmatrix}$.  Compute a basis for each of the eigenspaces of this matrix, $\mathcal{S}_1$, $\mathcal{S}_5$, and $\mathcal{S}_9$.
 \end{problem}
 
-\emph{Problems \ref{prob:3x3fromKuttler1}-\ref{prob:3x3fromKuttler2}}
-
-Let $A=\begin{bmatrix} 9 & 2 & 8\\ 2 & -6 & -2\\ -8 & 2 & -5\end{bmatrix}$.  
+\begin{problem}\label{prob:3x3fromKuttler1} Let $$A=\begin{bmatrix} 9 & 2 & 8\\ 2 & -6 & -2\\ -8 & 2 & -5\end{bmatrix}$$
 
-\begin{problem}\label{prob:3x3fromKuttler1}
-Compute the eigenvalues of this matrix. 
+\begin{question}
+Compute the eigenvalues of $A$. 
 \begin{hint}
 One of the eigenvalues of $A$ is -3.
 \end{hint}
@@ -302,16 +302,18 @@ \section*{Practice Problems}
 
 (List your answers in an increasing order.)
 $$\lambda_1 = \answer{-3},\quad \lambda_2 = \answer{-1},\quad \lambda_3 = \answer{2}$$
-\end{problem}
+\end{question}
 
-\begin{problem}\label{prob:3x3fromKuttler2}
-Compute a basis for each of the eigenspaces of this matrix,  $\mathcal{S}_{\lambda_1}$, $\mathcal{S}_{\lambda_2}$, and $\mathcal{S}_{\lambda_3}$.
+\begin{question}
+    
+Compute a basis for each of the eigenspaces of $A$,  $\mathcal{S}_{\lambda_1}$, $\mathcal{S}_{\lambda_2}$, and $\mathcal{S}_{\lambda_3}$.
 
 Answer:
 A basis for $\mathcal{S}_{\lambda_1}$ is $\left\{\begin{bmatrix}1\\\answer{2}\\\answer{-2}\end{bmatrix}\right\}$, 
 a basis for $\mathcal{S}_{\lambda_2}$ is $\left\{\begin{bmatrix}-2\\\answer{-2}\\\answer{3}\end{bmatrix}\right\}$,
 
 and a basis for $\mathcal{S}_{\lambda_3}$ is $\left\{\begin{bmatrix}2\\\answer{1}\\\answer{-2}\end{bmatrix}\right\}$.
+\end{question}
 \end{problem}
 
 
@@ -321,12 +323,17 @@ \section*{Practice Problems}
 
 
 \begin{problem}\label{prob:eigtri}
-Prove Theorem \ref{th:eigtri}.  (HINT:  Proceed by induction on the dimension n.  For the inductive step, compute $\det(A-\lambda I)$ by expanding along the first column (or row) if $T$ is upper (lower) triangular.)
+Prove Theorem \ref{th:eigtri}.  
+\begin{hint}
+    What do we know about the determinant of a triangular matrix?
+\end{hint}
+
+%(HINT:  Proceed by induction on the dimension n.  For the inductive step, compute $\det(A-\lambda I)$ by expanding along the first column (or row) if $T$ is upper (lower) triangular.)
 \end{problem}
 
-\emph{Problems \ref{prob:eigvectorstransfr2_1}-\ref{prob:eigvectorstransfr2_3}}
+% \emph{Problems \ref{prob:eigvectorstransfr2_1}-\ref{prob:eigvectorstransfr2_3}}
 
-The following set of problems deals with geometric interpretation of eigenvalues and eigenvectors, as well as linear transformations of the plane.  Please use \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/EIG-0010/main}{Describing Eigenvalues and Eigenvectors Algebraically and Geometrically} and \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/LTR-0070/main}{Geometric Transformations of the Plane} for reference.
+% The following set of problems deals with geometric interpretation of eigenvalues and eigenvectors, as well as linear transformations of the plane.  Please use \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/EIG-0010/main}{Describing Eigenvalues and Eigenvectors Algebraically and Geometrically} and \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/LTR-0070/main}{Geometric Transformations of the Plane} for reference.
 
 \begin{problem}\label{prob:eigvectorstransfr2_1}
 Recall that a vertical stretch/compression of the plane is a linear transformation whose standard matrix is $$M_v=\begin{bmatrix}1&0\\0&k\end{bmatrix}$$
@@ -348,14 +355,17 @@ \section*{Practice Problems}
 
 Answer:  A basis for $\mathcal{S}$ is $\left\{\begin{bmatrix}1\\\answer{0}\end{bmatrix}\right\}$
 
-Sketch several vectors in the eigenspace and use geometry to explain why the eigenvectors you sketched make sense.
+What is the algebraic multiplicity of $\lambda$? $\answer{2}$
+
+What is the geometric multiplicity of $\lambda$? $\answer{1}$
+
 \end{problem}
 
 \begin{problem}\label{prob:rotmatrixrealeig2}
 Recall that a counterclockwise rotation of the plane through angle $\theta$ is a linear transformation whose standard matrix is $$M_{\theta}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$
 Verify that the eigenvalues of $M_{\theta}$ are
 $$\lambda=\cos\theta\pm\sqrt{\cos^2\theta-1}$$
-Explain why $\lambda$ is a real number if and only if $\theta$ is a multiple of $\pi$.  (Compare this to Practice Problem \ref{prob:rotmatrixrealeig1} of \href{https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro/EIG-0010/main}{Describing Eigenvalues and Eigenvectors Algebraically and Geometrically}.)
+Use algebra, then use geometry to explain why $\lambda$ is a real number if and only if $\theta$ is a multiple of $\pi$.  
 
 Suppose $\theta$ is a muliple of $\pi$.  Then the eigenspaces corresponding to the two eigenvalues are the same.  Which of the following describes the eigenspace?
 \begin{multipleChoice}
@@ -401,6 +411,15 @@ \section*{Practice Problems}
 
 \end{problem}
 
+\begin{problem}\label{prob:eigenspace1}
+In this exercise we will prove that the eigenvectors associated with an eigenvalue $\lambda$ of an $n \times n$ matrix $A$, together with the zero vector, form a subspace of $\RR^n$.  To do this, follow the outline below.
+
+\begin{enumerate}
+\item
+Let $\vec{x}$ and $\vec{y}$ be eigenvectors of $A$ associated with $\lambda$.  Show that $\vec{x}+\vec{y}$ is also an eigenvector of $A$ associated with $\lambda$.  (This shows that the set of eigenvectors of $A$ associated with $\lambda$ is closed under addition).
+\item Show that the set of eigenvectors of $A$ associated with $\lambda$ is closed under scalar multiplication.
+\end{enumerate}
+\end{problem}
 
 \section*{Exercise Source}
 Practice Problem \ref{prob:3x3fromKuttler1} is adopted from Problem 7.1.11 of Ken Kuttler's \href{https://open.umn.edu/opentextbooks/textbooks/a-first-course-in-linear-algebra-2017}{\it A First Course in Linear Algebra}. (CC-BY)

week12.tex

@@ -0,0 +1,17 @@
+\documentclass{xourse}
+
+\input{preamble.tex}
+
+ \title{Week 12}
+
+\begin{document}
+\begin{abstract}
+\end{abstract}
+\maketitle
+
+%\part{Week 12 assignments}
+\activity{EIG-0020/main}
+\activity{EIG-0040/main}
+\activity{EIG-0050/main}
+
+\end{document}