\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

For each of the following matrices $A$ decide whether the origin
is a sink, source or saddle for the linear map $v \mapsto Av$.  If
a saddle, then draw the stable and unstable manifolds(lines):

(a) $\left(\begin{array}{cc} 4 & 30\\1 & 3 \end{array} \right),$\
\ \ \ (b) $\left(\begin{array}{cc} 1 & \frac{1}{2}\\\frac{1}{4} &
\frac{3}{4}
\end{array} \right),$\ \ \ \ (c) $\left(\begin{array}{rc} -0.4 & 2.4\\
-0.4 & 1.6 \end{array} \right)$
\medskip

{\bf Answer}
\begin{description}
\item[(a)]
Eigenvalues: $(\lambda-4)(\lambda-3)-30=0$ i.e.
$\lambda^2-7\lambda-18=0$: $(\lambda-9)(\lambda+2)=0$. Thus
$\lambda=9,-2$: both outside unit circle. Therefore
\underline{(flip) source}.

\item[(b)]
Eigenvalues: $(\lambda-1)(\lambda-\frac{3}{4})-\frac{1}{8}=0$ i.e.
$\lambda^2-\frac{7}{4}\lambda+\frac{5}{8}=0$:

$(\lambda-\frac{5}{4})(\lambda-\frac{1}{2})=0$. Thus
$\lambda=\ds\frac{5}{4},\ds\frac{1}{2}$: \underline{saddle} (no
flip).

Eigenvectors:

$\lambda=\ds\frac{5}{4}\ \ \left(\begin{array}{rr} -\frac{1}{4} &
\frac{1}{2}\\ \frac{1}{4} & -\frac{1}{2}
\end{array}\right)\left(\begin{array}{c}x\\y
\end{array}\right)=\left(\begin{array}{c}0\\0 \end{array}\right):\ \
\left(\begin{array}{c}x\\y
\end{array}\right)=\left(\begin{array}{c}2\\1 \end{array}\right)$
say.

$\lambda=\ds\frac{1}{2}\ \ \left(\begin{array}{cc} \frac{1}{2} &
\frac{1}{2}\\ \frac{1}{4} & \frac{1}{4}
\end{array}\right)\left(\begin{array}{c}x\\y
\end{array}\right)=\left(\begin{array}{c}0\\0 \end{array}\right):
\ \ \left(\begin{array}{c}x\\y
\end{array}\right)=\left(\begin{array}{c}1\\-1 \end{array}\right)$
say.

\begin{center}
\epsfig{file=314-3-4.eps, width=50mm}
\end{center}

\item[(c)]
Eigenvalues: $(\lambda+0.4)(\lambda-1.6)+0.96=0$ i.e.
$\lambda^2-1.2\lambda+0.32=0:\ (\lambda-0.4)(\lambda-0.8)=0$.

Thus $\lambda=0.4,0.8$: both inside the unit circle. Therefore
\underline{sink} (no flip).
\end{description}

\end{document}
