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{\bf Question}

Let $A$ be the $3 \times 3$ matrix $\ds\frac{1}{2}
\left(\begin{array} {rrr} 1 & -2 & -3\\ 5 & -26 & -45\\ -3 & 18 &
31 \end{array} \right)$.  Sketch the dynamics of the linear system
$v \mapsto Av$ in $\bf R$, indicating stable and unstable
manifolds.
\medskip

{\bf Answer}

The eigenvalues of $kA$ are $k \times$ eigenvalues of $A$.

Eigenvalues of $\left(\begin{array} {rrr} 1 & -2 & -3\\ 5 & -26 &
-45\\ -3 & 18 & 31 \end{array} \right)$ are solutions $\lambda$ to

$(1-\lambda)[(\lambda-31)(\lambda+26)+45.18]+2[5(31-\lambda)-3.45]-3[5.18-3(\lambda+26)]=0$

i.e.
$(1-\lambda)[\lambda^2-5\lambda+4]+2[-5\lambda+20]-3[-3\lambda+12]=0$

i.e. $(\lambda-4)[(1-\lambda)(\lambda-1)-10+9]=0$ (do
\underline{not} multiply out the previous line!)

i.e. $(\lambda-4)[-\lambda^2+2\lambda-2]=0$, so $\lambda=4, 1 \pm
i$.

Hence eigenvalues of the matrix $A$ in the question are
$\lambda=\underline{2,\frac{1}{2}(1 \pm i)}$.

Therefore the origin is a \underline{hyperbolic saddle}.

Eigenvectors:

$\lambda=2:\ \ \left(\begin{array}{rrr}-3 & -2 & -3\\5 & -30 &
-45\\-3 & 18 & 27\end{array}\right) \left(\begin{array}{r}x\\y\\z
\end{array} \right)=\left(\begin{array}{r}0\\0\\0 \end{array}
\right)$, so $\left(\begin{array}{r}x\\y\\z \end{array}
\right)=\left(\begin{array}{r}0\\3\\-2 \end{array} \right)$ (or
any scalar multiple of it).

$\lambda=\frac{1}{2}(1+i):\ \ \left(\begin{array}{ccc}-i & -2 &
-3\\5 & -27-i & -45\\-3 & 18 & 30-i\end{array}\right)
\left(\begin{array}{r}x\\y\\z
\end{array} \right)=\left(\begin{array}{r}0\\0\\0 \end{array}
\right)$

i.e. $\left(\begin{array}{ccc}-i & -2 & -3\\0 & -27+9i &
-45+15i\\0 & 18-6i & 30-10i\end{array}\right)
\left(\begin{array}{r}x\\y\\z
\end{array} \right)=\left(\begin{array}{r}0\\0\\0 \end{array}
\right)$

from which we get $(9-3i)y+(15-5i)z=0$ i.e. $-3y=5z$; then
$-ix=\frac{1}{5}y$.  So a (complex) \underline{eigenvector} for
$\lambda=\frac{1}{2}(1+i)$ is \underline{($i,5,-3$)}.

Hence let $\xi=(0,5,-3),\ \eta=(1,0,0)$

Then $A\xi=\frac{1}{2}\xi-\frac{1}{2}\eta,\
A\eta=\frac{1}{2}\xi+\frac{1}{2}\eta$. Hence in the plane spanned
by $\xi,\ \eta$ the action of $A$ is to apply the $2 \times 2$
matrix

$\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2}\\\frac{1}{2} &
\frac{1}{2} \end{array} \right)=\ds\frac{1}{\sqrt 2}
\left(\begin{array}{rr} \cos\frac{\pi}{4} &
-\sin\frac{\pi}{4}\\\sin\frac{\pi}{4} & \cos\frac{\pi}{4}
\end{array} \right)$.

i.e. \underline{rotate} by $\ds\frac{\pi}{4}$ and
\underline{shrink} by factor $\ds\frac{1}{\sqrt 2}$.

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