\documentclass[12pt]{article}
\usepackage{colordvi,amsmath,amssymb,shading,epic,curves}
\textheight=9.5in
\textwidth=7in
\hoffset=-0.75in
\voffset=-1.1in
\def\ds{\displaystyle}
\renewcommand{\theequation}{7.\arabic{equation}}
\renewcommand{\thepage}{\arabic{page}/6}
\renewcommand{\thefigure}{7.\arabic{figure}}

\begin{document}

\noindent {\LARGE \Blue{\textbf{MAT 275 Lecture 7}}}
\parashade[.95]{sharpcorners}{\noindent\Red{TOPICS}
Matrix formulation of $y''+\gamma y+4y=0$. Eigenvalues,
eigenvectors. Application to solution of HODEs when roots of characteristic
equation are distinct.}
\noindent In Lecture 5 we have seen how the ODE
\begin{equation}\label{eq1}
\textshade{sharpcorners}{$y''+\gamma y'+4y=0$}
\end{equation}
can be reduced to the system
\begin{equation}\label{eq2}
\frac{d}{dt}\begin{bmatrix}y\\z\end{bmatrix}
=\begin{bmatrix}ry+z\\-(r^2+\gamma r+4)y-(\gamma+r)z\end{bmatrix}
\end{equation}
using the intermediate variable $z=y'-ry$.
In this Lecture we define and use tools from Linear Algebra in the solution of
(\ref{eq2}). This approach will in particular greatly simplify the analytic
solution process when a right-hand side is added in the ODE in (\ref{eq1}).
Introduce
\begin{equation}\label{eq3}
\Red{\textbf{Matrix, vector}}\quad
A = \begin{bmatrix}r&1\\-(r^2+\gamma r+4)&-(\gamma+r)\end{bmatrix}
= \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix},
\quad
Y = \begin{bmatrix}y\\z\end{bmatrix} = \begin{bmatrix}Y_1\\Y_2\end{bmatrix}.
\end{equation}
and
\[
\begin{array}{c}\Red{\textbf{Matrix-vector multiplication}}
\\\Red{\textbf{Linear combination}}\end{array}\quad
\textshade[.9]{sharpcorners}{$AY=
\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}
\begin{bmatrix}Y_1\\Y_2\end{bmatrix}
=\begin{bmatrix}a_{11}Y_1+a_{12}Y_2\\a_{21}Y_1+a_{22}Y_2\end{bmatrix}
=Y_1\begin{bmatrix}a_{11}\\a_{21}\end{bmatrix}
+Y_2\begin{bmatrix}a_{12}\\a_{22}\end{bmatrix}$}
\]
The middle equality tells us how to apply (multiply) a matrix to (with) a vector.
The right equality shows how to write the resulting vector as linear
combination of the columns of the matrix. The matrix-vector multiplication is
generalized to
\[
\Red{\textbf{Matrix-matrix multiplication}}\quad
\textshade[.9]{sharpcorners}{$AB=A\left[\begin{array}{@{}c|c@{}}b_1&b_2\end{array}
\right]=\left[\begin{array}{@{}c|c@{}}Ab_1&Ab_2\end{array}\right]$}
\]
i.e., the matrix $AB$ is obtained by applying $A$ to each column of $B$
(the vertical bar $|$ emphasizes the matrix \textbf{partitioning}).\\[5pt]
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.1}}} \quad
$\ds \begin{bmatrix}3&2\\1&4\end{bmatrix}
\begin{bmatrix}1\\-2\end{bmatrix}=
\begin{bmatrix}-1\\-7\end{bmatrix},
\quad
\begin{bmatrix}3&2\\1&4\end{bmatrix}
\begin{bmatrix}1&2\\-2&3\end{bmatrix}
=\begin{bmatrix}-1&12\\-7&14\end{bmatrix}$.\\[10pt]
\noindent
\Blue{\textbf{We now assume that $r=r_1$ is a root of the characteristic equation
$r^2+\gamma r+4=0$}}. We recall that the two roots $r_1$ and $r_2$ are such that
$r_1+r_2=-\gamma$ and $r_1r_2=4$ (see Problem 6.6(a)). Then $A$ from (\ref{eq3})
reduces to
\begin{equation}\label{eq4}
 A=\begin{bmatrix}r_1&1\\0&r_2\end{bmatrix}.
\end{equation}
In this case $A$ is \Red{\textbf{upper triangular}}. Note that
\begin{equation}\label{eq5}
A\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}r_1\\0\end{bmatrix}
=r_1\begin{bmatrix}1\\0\end{bmatrix}
\quad\text{and}\quad
A\begin{bmatrix}1\\r_2-r_1\end{bmatrix}=\begin{bmatrix}r_1+(r_2-r_1)\\r_2(r_2-r_1)
\end{bmatrix}=r_2\begin{bmatrix}1\\r_2-r_1\end{bmatrix}.
\end{equation}
A \textbf{nonzero} vector $Y$ (i.e., at least one component is non-zero)
such that
\[ \Red{\textbf{E-vector $Y$, e-value $r$}}
\quad\textshade[.9]{sharpcorners}{$AY=rY$} \]
is called an \Red{\textbf{eigenvector of $A$ associated
to the eigenvalue $r$}}.
\newpage
\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.2}}} \quad
For the matrix $A$ from (\ref{eq4}):\\\vspace*{-1.5in}

\begin{minipage}{4in}
\begin{eqnarray*}
&&\begin{bmatrix}1\\0\end{bmatrix}
 \text{ is an eigenvector associated to eigenvalue } r_1 \\
&&\begin{bmatrix}1\\r_2-r_1\end{bmatrix}
 \text{ is an eigenvector associated to eigenvalue } r_2
\end{eqnarray*}
\end{minipage}
\quad
\begin{minipage}[t]{2in}
\setlength{\unitlength}{1in}
\begin{picture}(2,2)(-1,.5)
\put(-.5,0){\line(1,0){2}}\put(0,-.4){\line(0,1){1.5}}
\put(1,.5){\makebox(0,0)[b]{\footnotesize
$\begin{bmatrix}1\\r_2-r_1\end{bmatrix}$}}
\put(.5,-.05){\makebox(0,0)[t]{\footnotesize
$\begin{bmatrix}1\\0\end{bmatrix}$}}
\put(1,-.05){\makebox(0,0)[t]{1}}
\put(-.05,.5){\makebox(0,0)[r]{$r_2-r_1$}}
\dottedline{.05}(0,.5)(1,.5)(1,0)
\thicklines
\put(0,0){\Red{\vector(1,0){1}}}\put(0,0){\Red{\vector(2,1){1}}}
\end{picture}
\end{minipage}
\vspace*{.5in}

\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.3}}} \quad
Let $A=\begin{bmatrix}2&2\\2&-1\end{bmatrix}$. Then
\begin{alignat*}{2}
& A\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}6\\3\end{bmatrix}
=3\begin{bmatrix}2\\1\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}2\\1\end{bmatrix}\text{ is an eigenvector of $A$ associated to
eigenvalue } r=3 \\[3pt]
& A\begin{bmatrix}1\\-2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}
=(-2)\begin{bmatrix}1\\-2\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}1\\-2\end{bmatrix}\text{ is an eigenvector of $A$ associated to
eigenvalue } r=-2 \\[3pt]
& A\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}6\\0\end{bmatrix}
\ne r\begin{bmatrix}1\\2\end{bmatrix}\text{ for any }
r&&\quad\Longrightarrow\quad
\begin{bmatrix}1\\2\end{bmatrix}\text{ is \textbf{NOT} an eigenvector of } A.
\end{alignat*}
\begin{center}
\setlength{\unitlength}{.3in}
\begin{picture}(10,8.5)(-3,-3.5)
\multiput(-3,-3)(1,0){11}{\dottedline[\tiny .]{.2}(0,0)(0,8)}
\multiput(-3,-3)(0,1){9}{\dottedline[\tiny .]{.2}(0,0)(10,0)}
\put(1.1,-2){\makebox(0,0)[l]{\footnotesize
$\begin{bmatrix}1\\-2\end{bmatrix}$}}
\put(-1,2){\makebox(0,0)[rt]{\footnotesize
$A\begin{bmatrix}1\\-2\end{bmatrix}$}}
\put(1,2){\makebox(0,0)[rb]{\footnotesize
$\begin{bmatrix}1\\2\end{bmatrix}$}}
\put(4,-.1){\makebox(0,0)[t]{\footnotesize
$A\begin{bmatrix}1\\2\end{bmatrix}$}}
\put(2,1){\makebox(0,0)[br]{\footnotesize
$\begin{bmatrix}2\\1\end{bmatrix}$}}
\put(5,2.5){\makebox(0,0)[lt]{\footnotesize
$A\begin{bmatrix}2\\1\end{bmatrix}$}}
\put(0,0){\Blue{\vector(1,2){1}}}\put(0,0){\Blue{\vector(1,0){6}}}
\put(0,0){\vector(2,1){6}}\put(0,0){\vector(-1,2){2}}
\thicklines
\put(0,0){\Red{\vector(2,1){2}}}\put(0,0){\Red{\vector(1,-2){1}}}
\end{picture}
\end{center}

\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.4}}} \quad
Let $A=\begin{bmatrix}3&5\\-1&-1\end{bmatrix}$. Then
\begin{alignat*}{2}
& A\begin{bmatrix}5\\-2+i\end{bmatrix}
=\begin{bmatrix}5+5i\\-3-i\end{bmatrix}
=(1+i)\begin{bmatrix}5\\-2+i\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}5\\-2+i\end{bmatrix}\text{ e-vector associated to e-value }
r=1+i \\[3pt]
& A\begin{bmatrix}5\\-2-i\end{bmatrix}
=\begin{bmatrix}5-5i\\-3+i\end{bmatrix}
=(1-i)\begin{bmatrix}5\\-2-i\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}5\\-2-i\end{bmatrix}\text{ e-vector associated to e-value }
 r=1-i \\[3pt]
& A\begin{bmatrix}1\\-2\end{bmatrix}
=\begin{bmatrix}-7\\1\end{bmatrix}
\ne r\begin{bmatrix}1\\-2\end{bmatrix}\text{ for any }
r&&\quad\Longrightarrow\quad
\begin{bmatrix}1\\-2\end{bmatrix}\text{ is \textbf{NOT} an eigenvector of } A.
\end{alignat*}
\newpage
\noindent
\fbox{\textbf{\Red{Case $r_2\ne r_1$}}} The vectors
$\begin{bmatrix}1\\0\end{bmatrix}$ and
$\begin{bmatrix}1\\r_2-r_1\end{bmatrix}$ are distinct and ``point'' in
different directions.\\ Define
$X = \begin{bmatrix}1&1\\0&r_2-r_1\end{bmatrix}$. Note that
\begin{equation}\label{eq6}
X\begin{bmatrix}a\\b\end{bmatrix}=\mathbf{0}
\Leftrightarrow
\begin{bmatrix}a+b\\(r_2-r_1)b\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\Leftrightarrow
a=b=0
\Leftrightarrow
\begin{bmatrix}a\\b\end{bmatrix}=\mathbf{0}.
\end{equation}
Then
\begin{eqnarray*}
 AX &=& \left[\begin{array}{@{}cc@{}}A\begin{bmatrix}1\\0\end{bmatrix}
&A\begin{bmatrix}1\\r_2-r_1\end{bmatrix}\end{array}\right]\\
&=& \left[\begin{array}{@{}cc@{}}r_1\begin{bmatrix}1\\0\end{bmatrix}
&r_2\begin{bmatrix}1\\r_2-r_1\end{bmatrix}\end{array}\right]\\
&=& \left[\begin{array}{@{}cc@{}}r_1X\begin{bmatrix}1\\0\end{bmatrix}
&r_2X\begin{bmatrix}0\\1\end{bmatrix}\end{array}\right]\\
&=& \left[\begin{array}{@{}cc@{}}X\begin{bmatrix}r_1\\0\end{bmatrix}
&X\begin{bmatrix}0\\r_2\end{bmatrix}\end{array}\right]\\
&=& X\begin{bmatrix}r_1&0\\0&r_2\end{bmatrix}\\&=&XD.
\end{eqnarray*}
We now use these relations to solve the IVP (\ref{eq1}). Introduce a vector
\[ Z=Z(t)=\begin{bmatrix}z_1(t)\\z_2(t)\end{bmatrix}\]
such that $Y(t)=XZ(t)$. Then
\[ Y'=\left\{\begin{array}{ll}XZ' &\text{(since $X$ is a constant matrix)}\\
   AY=AXZ=XDZ & \text{(from (\ref{eq2}) and the above property)}
   \end{array}\right. \]
so that
\begin{eqnarray}
XZ'=XDZ &\quad\Leftrightarrow\quad & X(Z'-DZ)=\mathbf{0}\nonumber\\
        &\quad\Leftrightarrow\quad & Z'-DZ=\mathbf{0}
        \quad\text{(because of (\ref{eq6}))}\nonumber\\
        &\quad\Leftrightarrow\quad &
         \begin{bmatrix}z'_1\\z'_2\end{bmatrix}
         =\begin{bmatrix}r_1&0\\0&r_2\end{bmatrix}
        \begin{bmatrix}z_1\\z_2\end{bmatrix}\nonumber\\
        &\quad\Leftrightarrow\quad &
          \left\{\begin{array}{l}z'_1=r_1z_1\\
                       z'_2=r_2z_2\end{array}\right.
        \quad\text{(solve 2 \textbf{independent} first-order ODEs)}
        \nonumber\\
        &\quad\Leftrightarrow\quad &
        \left\{\begin{array}{l}z_1(t)=C_1e^{r_1t}\\
          z_2(t)=C_2e^{r_2t}\end{array}\right.\nonumber\\
        &\quad\Leftrightarrow\quad & Z(t)=\begin{bmatrix}C_1e^{r_1t}\\
                         C_2e^{r_2t}\end{bmatrix}\nonumber\\
        &\quad\Leftrightarrow\quad &
        \textshade[.9]{sharpcorners}{$Y(t)$}=X\begin{bmatrix}C_1e^{r_1t}\\
                          C_2e^{r_2t}\end{bmatrix}
        \textshade[.9]{sharpcorners}{$
        =C_1e^{\Blue{r_1}t}\Red{\begin{bmatrix}1\\0\end{bmatrix}}
   +C_2e^{\Blue{r_2}t}\Red{\begin{bmatrix}1\\r_2-r_1\end{bmatrix}}$}.
\label{eq7}
%        =\left[\begin{array}{@{}c|c@{}}
%        e^{r_1t}\begin{bmatrix}1\\0\end{bmatrix}&
%        e^{r_2t}\begin{bmatrix}1\\r_2-r_1\end{bmatrix}\end{array}\right]
%        \begin{bmatrix}C_1\\C_2\end{bmatrix}.
\end{eqnarray}
The first component of $Y(t)$ is $y(t)=C_1e^{r_1t}+C_2e^{r_2t}$ which is
identical to what was obtained in Lecture 6.\\[10pt]
\vspace*{-10pt}

\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.5}}} \quad
Solve \textshade[.95]{sharpcorners}{$\ds\frac{d}{dt}
\begin{bmatrix}y\\z\end{bmatrix}
=\begin{bmatrix}-1&1\\0&-4\end{bmatrix}
\begin{bmatrix}y\\z\end{bmatrix}$
with $\begin{bmatrix}y(0)\\z(0)\end{bmatrix}=
\begin{bmatrix}1\\3\end{bmatrix}$.}\\[3pt]
The matrix is of the form (\ref{eq4}) with $r_1=-1$ and $r_2=-4$.
It has corresponding eigenvectors $\begin{bmatrix}1\\0\end{bmatrix}$ and
$\begin{bmatrix}1\\-3\end{bmatrix}$, respectively. The general solution of
$Y'=AY$ is then
\begin{equation}\label{eq8}
 Y(t)=C_1 e^{-t}\begin{bmatrix}1\\0\end{bmatrix}
+C_2e^{-4t}\begin{bmatrix}1\\-3\end{bmatrix}. \end{equation}
Then $\ds\quad Y(0)=C_1\begin{bmatrix}1\\0\end{bmatrix}
+C_2\begin{bmatrix}1\\-3\end{bmatrix}=
\begin{bmatrix}1\\3\end{bmatrix}
\quad\Longrightarrow\quad
\left\{\begin{array}{l}C_1+C_2=1\\-3C_2=3\end{array}\right.
\quad\Longrightarrow\quad
\left\{\begin{array}{l}C_1=2\\C_2=-1
\end{array}\right.$\\[3pt]
i.e., $Y(t)=2e^{-t}\begin{bmatrix}1\\0\end{bmatrix}
-e^{-4t}\begin{bmatrix}1\\-3\end{bmatrix}$, or
\textshade[.95]{sharpcorners}{$\begin{array}{rcl}y(t)&=&2e^{-t}-e^{-4t}
\\z(t)&=&3e^{-4t}\end{array}$}. We verify that $z=y'-r_1y=y'+y$.\\[20pt]
\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.6}}} \quad
Solve \textshade[.95]{sharpcorners}{$\ds\frac{d}{dt}
\begin{bmatrix}y\\u\end{bmatrix}
=\begin{bmatrix}-2&2\\1&-3\end{bmatrix}
\begin{bmatrix}y\\u\end{bmatrix}$
with $\begin{bmatrix}y(0)\\u(0)\end{bmatrix}=
\begin{bmatrix}1\\2\end{bmatrix}$.}\\[3pt]
The matrix $A=\begin{bmatrix}-2&2\\1&-3\end{bmatrix}$ does not have the form
(\ref{eq4}). However
\[ P=\begin{bmatrix}1&0\\-1&2\end{bmatrix} \quad\text{and}\quad
Q=\begin{bmatrix}1&0\\\frac{1}{2}&\frac{1}{2}\end{bmatrix} \quad\text{satisfy}
\quad PA=\begin{bmatrix}-1&1\\0&-4\end{bmatrix}P
\quad\text{and}\quad
PQ=QP=\begin{bmatrix}1&0\\0&1\end{bmatrix}.
\]
Then the quantity $Y=P\begin{bmatrix}y\\u\end{bmatrix}$ is such that
\[ Y'=P\frac{d}{dt}\begin{bmatrix}y\\u\end{bmatrix}
=\Fuchsia{PA}\begin{bmatrix}y\\u\end{bmatrix}
=\Fuchsia{\begin{bmatrix}-1&1\\0&-4\end{bmatrix}P}\begin{bmatrix}y\\u\end{bmatrix}
=\begin{bmatrix}-1&1\\0&-4\end{bmatrix}Y. \]
Using the form (\ref{eq8}) of the solution $Y(t)$ we obtain
\begin{equation}\label{eq9}
\begin{bmatrix}y(t)\\u(t)\end{bmatrix}
=QP\begin{bmatrix}y(t)\\u(t)\end{bmatrix}=QY(t)=C_1 e^{-t}
Q\begin{bmatrix}1\\0\end{bmatrix}+C_2e^{-4t}Q\begin{bmatrix}1\\-3\end{bmatrix}
=C_1 e^{-t}\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}
+C_2e^{-4t}\begin{bmatrix}1\\-1\end{bmatrix}.
\end{equation}
Note that
\begin{alignat*}{2}
& A\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}
=\begin{bmatrix}-1\\-\frac{1}{2}\end{bmatrix}
=(-1)\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}\text{ e-vector \textbf{of $A$}
associated to e-value } r=-1 \\[3pt]
& A\begin{bmatrix}1\\-1\end{bmatrix}
=\begin{bmatrix}-4\\4\end{bmatrix}
=(-4)\begin{bmatrix}1\\-1\end{bmatrix}&&\quad\Longrightarrow\quad
\begin{bmatrix}1\\-1\end{bmatrix}\text{ e-vector \textbf{of $A$}
associated to e-value } r=-4.
\end{alignat*}
Then $\ds\quad \begin{bmatrix}y(0)\\u(0)\end{bmatrix}
=C_1\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}
+C_2\begin{bmatrix}1\\-1\end{bmatrix}=
\begin{bmatrix}1\\2\end{bmatrix}
\quad\Longrightarrow\quad
\left\{\begin{array}{l}C_1+C_2=1\\\frac{1}{2}C_1-C_2=2\end{array}\right.
\quad\Longrightarrow\quad
\left\{\begin{array}{l}C_1=2\\C_2=-1\end{array}\right.$\\[3pt]
i.e., $\begin{bmatrix}y(t)\\u(t)\end{bmatrix}
=2e^{3t}\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}
-e^{-2t}\begin{bmatrix}1\\-1\end{bmatrix}$, or
\textshade[.95]{sharpcorners}{$\begin{array}{rcl}y(t)&=&2e^{-t}-e^{-4t}
\\u(t)&=&e^{-t}+e^{-4t}\end{array}$}\\[5pt]
Note that $y(t)$ is identical to the solution obtained in Example 7.5.

\newpage
\noindent
Example 7.5 suggests the following result (see Problem \ref{pb7}):

\noindent
\parashade[1]{sharpcorners}{\noindent
If $A$ is a $2\times 2$ matrix with distinct eigenvalues
the general solution of $Y'=AY$ is a linear combination
\Red{
\[ Y(t)=C_1\; e^{\text{($1^{\text{\footnotesize st}}$ eigenvalue of $A$)}\; t}
\begin{bmatrix}\text{$1^{\text{\footnotesize st}}$}\\\text{eigenvector}\\
\text{of $A$}\end{bmatrix}
+C_2\; e^{\text{($2^{\text{\footnotesize nd}}$ eigenvalue of $A$)}\; t}
\begin{bmatrix}\text{$2^{\text{\footnotesize nd}}$}\\\text{eigenvector}\\
\text{of $A$}\end{bmatrix}.\]
}
The constants $C_1$ and $C_2$ are determined from the initial conditions.}


\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.7}}} \quad
Solve \textshade[.95]{sharpcorners}{$\ds\frac{d}{dt}
\begin{bmatrix}y\\z\end{bmatrix}
=\begin{bmatrix}2&2\\2&-1\end{bmatrix}
\begin{bmatrix}y\\z\end{bmatrix}$
with $\begin{bmatrix}y(0)\\z(0)\end{bmatrix}=
\begin{bmatrix}1\\3\end{bmatrix}$.}\\[3pt]
Using eigenvalues and eigenvectors from Example 7.3 the general solution of $Y'=AY$ is
\[ Y(t)=C_1 e^{3t}\begin{bmatrix}2\\1\end{bmatrix}
+C_2e^{-2t}\begin{bmatrix}1\\-2\end{bmatrix}. \]
Then $\ds\quad Y(0)=C_1\begin{bmatrix}2\\1\end{bmatrix}
+C_2\begin{bmatrix}1\\-2\end{bmatrix}=
\begin{bmatrix}1\\3\end{bmatrix}
\quad\Longrightarrow\quad
\left\{\begin{array}{l}2C_1+C_2=1\\C_1-2C_2=3\end{array}\right.
\quad\Longrightarrow\quad
\left\{\begin{array}{l}C_1=1\\C_2=-1\end{array}\right.$\\[3pt]
i.e., $Y(t)=e^{3t}\begin{bmatrix}2\\1\end{bmatrix}
-e^{-2t}\begin{bmatrix}1\\-2\end{bmatrix}$, or
\textshade[.95]{sharpcorners}{$\begin{array}{rcl}y(t)&=&2e^{3t}-e^{-2t}
\\z(t)&=&e^{3t}+2e^{-2t}\end{array}$}\\[20pt]

\noindent
\textshade[.7]{sharpcorners}{\textbf{\White{EXAMPLE 7.8}}} \quad
Solve \textshade[.95]{sharpcorners}{$\ds\frac{d}{dt}
\begin{bmatrix}y\\z\end{bmatrix}
=\begin{bmatrix}3&5\\-1&-1\end{bmatrix}
\begin{bmatrix}y\\z\end{bmatrix}$
with $\begin{bmatrix}y(0)\\z(0)\end{bmatrix}=
\begin{bmatrix}0\\2\end{bmatrix}$.}\\[3pt]
Using eigenvalues and eigenvectors from Example 7.4 the general solution of
 $Y'=AY$ is
\[ Y(t)=C_1 e^{(1+i)t}\begin{bmatrix}5\\-2+i\end{bmatrix}
+C_2e^{(1-i)t}\begin{bmatrix}5\\-2-i\end{bmatrix}. \]
Then $\ds\quad Y(0)=C_1\begin{bmatrix}5\\-2+i\end{bmatrix}
+C_2\begin{bmatrix}5\\-2-i\end{bmatrix}=
\begin{bmatrix}0\\2\end{bmatrix}
\quad\Longrightarrow\quad
\left\{\begin{array}{l}5C_1+5C_2=0\\
(-2+i)C_1+(-2-i)C_2=2\end{array}\right.$\\[3pt]
yields $C_1=-i$, $C_2=i$ (Verify!),\\ i.e.,
 $Y(t)=-ie^{(1+i)t}\begin{bmatrix}5\\-2+i\end{bmatrix}
+ie^{(1-i)t}\begin{bmatrix}5\\-2-i\end{bmatrix}
=e^t\begin{bmatrix}-5ie^{it}+5ie^{-it}\\
-i(-2+i)e^{it}+i(-2-i)e^{-it}\end{bmatrix}$,\\[5pt]
\hspace*{2in} or
\textshade[.95]{sharpcorners}{$\begin{array}{rcl}y(t)&=&10e^t\sin t
\\z(t)&=&2e^t\cos t-4e^t\sin t\end{array}$}

\newpage
\noindent
\fbox{\textbf{\Red{Problems}}}
\begin{enumerate}
\item Consider the matrices $A=\begin{bmatrix}3&2\\1&4\end{bmatrix}$,
$A=\begin{bmatrix}3&2\\1&4\end{bmatrix}$, and
$A=\begin{bmatrix}3&2\\1&4\end{bmatrix}$, and the vector
$Y=\begin{bmatrix}3\\-2\end{bmatrix}$.
\begin{enumerate}
\item Determine the quantities $AY$, $BY$, $E=AB$, $F=BC$, $BA$. Is $AB=BA$?
\item Compare $EC$ and $AF$. What rule can you infer from this result?
\end{enumerate}
\item Let $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and
$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and
$Y=\begin{bmatrix}Y_1\\Y_2\end{bmatrix}$. Verify that $IY=Y$ and $AI=IA=A$.
The matrix $I$ is called the \Red{\textbf{Identity}} matrix.
\item Show that if $Y$ is an eigenvector of $A$ then $\alpha Y$
is also an eigenvector for any $\alpha\ne0$.
\item \label{pb3} Let $A=\begin{bmatrix}-2&1\\1&-2\end{bmatrix}$.
\begin{enumerate}
\item Verify that $\begin{bmatrix}1\\1\end{bmatrix}$ and
$\begin{bmatrix}1\\-1\end{bmatrix}$ are eigenvectors. What are the corresponding
eigenvalues?
\item Solve $Y'=AY$ with $Y(0)=\begin{bmatrix}1\\2\end{bmatrix}$.
\item Implement the IVP in \textsc{Matlab} and compare numerical and exact
solutions.
\end{enumerate}
\item Repeat Problem \ref{pb3} for the matrix
$A=\begin{bmatrix}3&2\\1&4\end{bmatrix}$ with eigenvectors
$\begin{bmatrix}2\\-1\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$.
\item Reduce the IVP from Problem \ref{pb3}(b) to a second-order linear IVP
in $Y_1$ (i.e., eliminate $Y_2$ from the two first-order ODEs). What is the
characteristic equation of the resulting ODE? How do the roots compare to the
eigenvalues of the matrix? (do Problem \ref{pb3}(a) first).
\item Consider the matrix $A$ from (\ref{eq3}) with $\gamma=5$. We consider
the three choices $r=-1,0,1$.
\begin{enumerate}
\item Write down the three IVPs resulting from the these choices (do not forget
the ICs!)
\item Identify the following vectors as eigenvector of one of the system matrices:
\[ \begin{bmatrix}1\\-5\end{bmatrix},\quad
   \begin{bmatrix}1\\-4\end{bmatrix},\quad
   \begin{bmatrix}1\\-3\end{bmatrix},\quad
   \begin{bmatrix}1\\-2\end{bmatrix},\quad
   \begin{bmatrix}1\\-1\end{bmatrix},\quad
   \begin{bmatrix}1\\0\end{bmatrix}. \]
What are the corresponding eigenvalues?
\item Solve the three IVPs and verify that the first component of the solution is the same for all three cases.
\end{enumerate}
\item \label{pb7}
Assume that $AP=PB$ with $B=\begin{bmatrix}r_1&1\\&r_2\end{bmatrix}$, $r_2\ne r_1$,
and $PQ=QP=I$ ($P$ is said to be \Red{invertible} or \Red{nonsingular}.
The matrices $A$ and $B$ are \Red{similar}).
\begin{enumerate}
\item Show that the vectors $P\begin{bmatrix}1\\0\end{bmatrix}$ and
$P\begin{bmatrix}1\\r_2-r_1\end{bmatrix}$ are eigenvectors of $A$. What are the
associated eigenvalues?
\item Let $\widetilde{Y}$ be the general solution of
$\widetilde{Y}'=A\widetilde{Y}$. Show that $Y=Q\widetilde{Y}$ is solution of
$Y'=BY$.
\item Determine the general solution $\widetilde{Y}(t)$.
\item Is it always possible to find matrices $P$, $Q$, and $B$ with the
above properties?
\end{enumerate}
\end{enumerate}

\end{document}