\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Find (a) $\ds\lim_{n \to \infty} \left(\begin{array}{rr} 4.5 &
8\\-2 & -3.5 \end{array} \right)^n\left(\begin{array} {c} 6\\ 9
\end{array} \right)$,\ \ \ \ (b) $\ds\lim_{n \to \infty} \left(\begin{array}{rr}
\frac{3}{2} & 1\\-\frac{1}{2} & 0 \end{array}
\right)^n\left(\begin{array} {c} 6\\ 9
\end{array} \right)$
\medskip

{\bf Answer}
\begin{description}
\item[(a)]
$A=\left(\begin{array} {rr} 4.5 & 8\\ -2 & -3.5 \end{array}
\right)$ has eigenvalues: $(\lambda-4.5)(\lambda+3.5)+16=0$

i.e. $\lambda^2-\lambda=0.25=0$ i.e.
$\lambda=\frac{1}{2},\frac{1}{2}$.

These are inside the unit circle, so the origin is a
\underline{sink}. This means \underline{$A^nv \rightarrow (0,0)$}
for \underline{every} vector $v \in {\bf R}^2$ (not just (6,9)).

\item[(b)]
$A=\left(\begin{array} {rr} \frac{3}{2} & 1\\ -\frac{1}{2} & 0
\end{array} \right)$ has eigenvalues:
$(\lambda-\frac{3}{2})\lambda+\frac{1}{2}=0$

i.e. $\lambda^2-\frac{3}{2}\lambda+\frac{1}{2}=0:\
(\lambda-1)(\lambda-\frac{1}{2})=0$. Hence
$\lambda=1,\ds\frac{1}{2}$.

This means (0,0) is a non-hyperbolic fixed point.

Eigenvectors:

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

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

\begin{center}
\epsfig{file=314-3-5.eps, width=50mm}
\end{center}

Thus under the action of $A$, every point on the line
$y=-\frac{1}{2}x$ remains fixed (there is a whole line of fixed
points), while every vector in the direction $y=-x$ is shrunk by a
factor $\ds\frac{1}{2}$.  Hence the point (6,9) will be attracted
to the point where the line through (6,9) with slope -1 meets the
line $y=\frac{-x}{2}$. Easy to check this point is (30,-15).
\end{description}
\end{document}








