\documentclass[12pt]{article}
\usepackage{latexsym}
%\input psfig.sty
\input spec_frac
\begin{document}
\title{314 '03-QUIZ 11}
\author{Name:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ }
\maketitle
\section{$<12pts>$}
Compute $e^{A}$ if $A$ is the matrix
\[ A = 
\left[
\begin{array}{rr}
3 & 4 \\
\ -2  & \ -3
\end{array}
\right]
\]
%Let $A$ be a $2\times 2$ matrix and let $p(\lambda) = \lambda^2 + b \lambda +c$
%be the characteristic polynomial of A. Show that 
%\[ b = -Tr(A) = -\left(a_{11}+a_{22}\right) \ \ ,\ \ c = det(A) = a_{11}a_{22}-a_{12}a_{21}\ 
%\ \mbox{with}\ \ A = 
%\left[
%\begin{array}{rr}
%a_{11} & a_{12} \\
%a_{21} & a_{22}
%\end{array}
%\right]
%\]
%where $Tr(A)$ is the trace of the matrix $A$. \\
{\bf Solution}\\
We need to find eigenvalues and eigenvectors of $A= V D V^{-1}$:
%The characteristic polynomial is by definition:
\[ det\left(A-\lambda I\right) =
\left| \begin{array}{rr}
3-\lambda & 4 \\
-2 & -3-\lambda 
\end{array} \right| = 
\lambda^2 - 1 = 0 \rightarrow \lambda = \pm 1 \]
%in the form desired.
\begin{enumerate}
\item $\lambda_1 = 1$:
\[ \left[\begin{array}{rr}
2 & 4 \\ -2 & -4 
\end{array} \right] \rightarrow {\bf v}_1 = 
 \left[\begin{array}{r}
-2 \\ 1 
\end{array} \right] 
\]
\item $\lambda_2 = -1$:
\[ \left[\begin{array}{rr}
4 & 4 \\ -2 & -2 
\end{array} \right] \rightarrow {\bf v}_2 =
 \left[\begin{array}{r}
1 \\ -1 
\end{array} \right] 
\]
Then
\[ V = \left[\begin{array}{rr} -2 & 1 \\ 1 & -1 \end{array} \right]\ ,\ 
   D = \left[\begin{array}{rr} e^1 & 0 \\ 0 & e^{-1} \end{array} \right]\ ,\
   V^{-1} = \left[\begin{array}{rr} -1 & -1 \\ -1 & -2
\end{array} \right]
 \]
so
\begin{eqnarray*} 
e^{A} & = & V e^D V^{-1} =
\left[\begin{array}{rr} -2 & 1 \\ 1 & -1 \end{array} \right]
\left[\begin{array}{rr} e^{1} & 0 \\ 0 & e^{-1} \end{array} \right]
\left[\begin{array}{rr} -1 & -1 \\ -1 & -2
\end{array} \right]  \\
& = &
\left[\begin{array}{rr} e^{1} & e^{-1} \\ -2e^{1} & -e^{-1} \end{array} \right]
\left[\begin{array}{rr} -1 & -1 \\ 2 & 1
\end{array} \right]
= 
\left[\begin{array}{rr}
-e^{1}+2e^{-1} & -e^{1}+e^{-1} \\
2e^{1} -2e^{-1}& 2e^{1} -e^{-1}
\end{array} \right]
\end{eqnarray*}
\end{enumerate}


\newpage

\section{$<13pts>$}
Solve the initial value problem:
\begin{eqnarray*}
y_1^\prime & = & \ \ \ 3y_1 +y_2  \ ,\  y_1(0) = 1\\
y_2^\prime & = & -4y_1 -2y_2\ ,\  y_2(0) = 2
\end{eqnarray*}

{\bf Solution}\\
\[
\sfrac{d{\bf y}}{dt} =  A {\bf y} :=
\left[
\begin{array}{rrr}
3  & 1 \\  -4  & -2
\end{array}
\right]\left[
\begin{array}{r}
y_1 \\ y_2 
\end{array}
\right] \rightarrow {\bf y} = e^{At} {\bf y}_0\ 
= Ve^{Dt}V^{-1} {\bf y}_0,\]
\begin{eqnarray*}
A= VDV^{-1}
\ ,\ V = \left[{\bf v}_1, {\bf v}_2
\right] &,  & D = \left[
\begin{array}{rr}
\lambda_1 & 0 \\ 0 & \lambda_2
\end{array}
\right] \\
A {\bf v}_i = \lambda_i {\bf v}_i \rightarrow
\left|
\begin{array}{rrr}
3-\lambda & 1 \\ -4 & -2-\lambda
\end{array}
\right| & = & 
\left(3-\lambda\right)\left(-2-\lambda\right)
- \left(1 \right)\left(-4\right) = 0 \\
\lambda^2 - \lambda -2 = 0
& \rightarrow & 
\left(\lambda-2\right)\left(\lambda+1\right) = 0 \rightarrow
\lambda = 2, -1
\end{eqnarray*}
We find the eigenvectors
\begin{enumerate}
\item $\lambda_1 = 2$
\[
\begin{array}{r}
\ \\ 4
\end{array}
\left[
\begin{array}{rrr}
1 & 1 \\ -4 & -4
\end{array}
\right] \rightarrow
\left[
\begin{array}{rr}
1 & 1 \\ 0 & 0
\end{array}
\right]\rightarrow {\bf v}_1 = \left[
\begin{array}{r}
1 \\ -1
\end{array}
\right]
\]
\item $\lambda_2 = -1$
\[
\begin{array}{r}
\ \\ 1
\end{array}
\left[
\begin{array}{rrr}
4 & 1 \\ -4 & -1
\end{array}
\right] \rightarrow
\left[
\begin{array}{rr}
4 & 1 \\ 0 & 0
\end{array}
\right]\rightarrow {\bf v}_2 = \left[
\begin{array}{r}
1 \\ -4
\end{array}
\right]
\]
\end{enumerate}
Then
\[
V = \left[
\begin{array}{rr}
1 & 1 \\ -1 & -4 
\end{array}
\right]\ ,\ V^{-1} =
\sfrac{-1}{3}
\left[
\begin{array}{rr}
-4 & -1 \\ 1 & 1 
\end{array}
\right] \ , \ 
e^{Dt} = \left[
\begin{array}{rr}
e^{2t} & 0 \\ 0 & e^{-t}
\end{array}
\right]
\]
so that
\begin{eqnarray*}
{\bf y}(t) & = &
\left(
\left[
\begin{array}{rr}
1 & 1 \\ -1 & -4 
\end{array}
\right]\left[
\begin{array}{rr}
e^{2t} & 0 \\ 0 & e^{-t}
\end{array}
\right]\right)\left(\sfrac{-1}{3}
\left[
\begin{array}{rr}
-4 & -1 \\ 1 & 1 
\end{array}
\right] 
\left[
\begin{array}{r}
1 \\ 2
\end{array}
\right] \right) \\
 & = & \left[
\begin{array}{rr}
e^{2t} & e^{-t} \\ -e^{2t} & -4e^{-t} 
\end{array}
\right] 
\left[
\begin{array}{r}
2 \\ -1
\end{array}
\right]\\
& = & \left[
\begin{array}{r}
2e^{2t} - e^{-t} \\
-2e^{5t}  +4e^{-t}
\end{array}
\right]
\end{eqnarray*}

\newpage
\begin{center}
{\bf \Large{314 '03-QUIZ 12}}
\end{center}

\section{$< 12pts >$}
Given the matrices $A$, $B$:
\[ A = \left[ 
\begin{array}{rrr}
2 & 2 \\ 2 & -1
\end{array}
\right] \ ,\  
B = \left[ 
\begin{array}{rrr}
2 & 0 \\ 0 & -1
\end{array}
\right]
\]
Is there a nonsingular matrix $S$ such that
\[ S^{-1} A S = B \ ? \]
If so find it. If not, explain why not.\\
{\bf Solution}\\
The problem asks for a similarity transformation between $A$ and $B$. This
is only possible if they have the same eigenvalues. Since $B$ is diagonal, its
eigenvalues are exhibited by its diagonal elements, i.e. they are $2\ ,\ -1$.
For $A$ they need to be computed:
\[ det(A-\lambda I) = \left| 
\begin{array}{rrr}
2-\lambda & 2 \\ 2 & -1-\lambda 
\end{array}
\right| = \lambda^2- \lambda -6 = 0 \rightarrow \lambda = 3\ ,\  -2
\]
Therefore the two matrices have different eigenvalues, and they are not
similar. No matrix $S$ with desired properties can be found.

\newpage

\section{$< 13pts >$}
A matrix $A$ is normal if $AA^H = A^H A$. Show that the matrix A is
normal and
find an orthogonal or unitary diagonalizing matrix $V$ such that
$V^H A V = D$ with $D$ diagonal,  
where
\[ A = \left[
\begin{array}{rr}
0 & 2+i  \\ -2+i & 0 
\end{array}
\right]
\]

{\bf Solution}\\
\[ A^H = \left[
\begin{array}{rrr}
0 & -2-i \\ 2-i & 0
\end{array}
\right]= -A \ \rightarrow \  A A^H = A^H A = - A^2
\]
so that $A$ is normal.

\[ A - \lambda I = \left| 
\begin{array}{rrr}
-\lambda & 2+i \\ -2+i & -\lambda 
\end{array}
\right| = \lambda^2 - (2+i)(-2+i)
 =  \lambda^2+5 = 0
\]

\begin{enumerate}

\item $\lambda_1 = i\sqrt{5}$
\begin{eqnarray*}
\begin{array}{r}
\ \\ 0 \\ \frac{1+2i}{\sqrt{5}}
\end{array} 
\left[
\begin{array}{rrr}
-i\sqrt{5} & 2+i  \\ -2+i & -i\sqrt{5}
\end{array}
\right] & \rightarrow &  
\left[
\begin{array}{rrr}
-i\sqrt{5} & 2+i \\ 0 & 0
\end{array}
\right] \rightarrow {\bf v}_1 = \left[
\begin{array}{r}
2+i \\ i\sqrt{5} 
\end{array}
\right]
\end{eqnarray*}

\item $\lambda_2 = \ -i\sqrt{5}$
\begin{eqnarray*}
\begin{array}{r}
\ \\ 0 \\ -\frac{1+2i}{\sqrt{5}}
\end{array} 
\left[
\begin{array}{rrr}
i\sqrt{5} & 2+i  \\ -2+i & i\sqrt{5}
\end{array}
\right] & \rightarrow &  
\left[
\begin{array}{rrr}
i\sqrt{5} & 2+i \\ 0 & 0
\end{array}
\right] \rightarrow {\bf v}_2 = \left[
\begin{array}{r}
2+i \\ -i\sqrt{5} 
\end{array}
\right]
\end{eqnarray*}
The two eigenvectors are orthogonal:
\[ {\bf v}_1^H {\bf v}_2 = (2-i)(2+i)+(i\sqrt{5})^2 = 0 \]
so, after dividing each by its magnitude we get a unitary
matrix:
\[ U = \left[ \frac{{\bf v}_1}{||{\bf v}_1||}\ ,\ 
              \frac{{\bf v}_2}{||{\bf v}_2||} \right] =
\left[
\begin{array}{rr}
\frac{2+i}{\sqrt{10}} & \frac{2+i}{\sqrt{10}} \\
\frac{i\sqrt{5}}{\sqrt{10}} & -\frac{i\sqrt{5}}{\sqrt{10}} 
\end{array} \right]
\]

\end{enumerate}
\end{document}
