Page 409 Problem 1

Page 409 Problem 1

$\left(\begin{array}{cc}
3 & -2\\
4 & -1\end{array}\right)$ gives $\lambda^{2}-(2)\lambda+(-3-(-8))=\lambda^{2}-2\lambda+5$

Solving for the roots we have $\lambda=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i$

Let's find the eigenvector for $\lambda=1+2i$

$\left(\begin{array}{cc}
2-2i & -2\\
4 & -2-2i\end{array}\right)$

As descried in the review, we should multiply the top row by $\frac{4}{2-2i}$
and see that we get the bottom row.

$-2\cdot\frac{4}{2-2i}=\frac{-4}{1-i}=\frac{-4}{1-i}\frac{1+i}{1+i}=-\frac{4}{2}(1+i)=-2-2i$.
Great!

That was the algorithmic way to go about it. More heuristically,

$\to\left(\begin{array}{cc}
1-i & -1\\
2 & -1-i\end{array}\right)$. Can see that should multiply by the conjugate $1+i$.

$\to\left(\begin{array}{cc}
(1-i) & (-1)\\
0 & 0\end{array}\right)$

Write $\xi=\left(\begin{array}{c}
\xi_{1}\\
\xi_{2}\end{array}\right)$

Let $\xi_{1}=\alpha$. Then $x_{2}=(1-i)\alpha$.

We conclude $\xi=\left(\begin{array}{c}
\alpha\\
(1-i)\alpha\end{array}\right)=\left(\begin{array}{c}
1\\
1-i\end{array}\right)\alpha$

So then a solution is $\xi e^{\lambda t}=e^{t}\left(\begin{array}{c}
1\\
1-i\end{array}\right)\left(\cos2t+i\sin2t\right)=e^{t}\left[\left(\begin{array}{c}
\cos2t\\
\cos2t+\sin2t\end{array}\right)+i\left(\begin{array}{c}
\sin2t\\
-\cos2t+\sin2t\end{array}\right)\right]$

Thus we have $x_{1}(t)=e^{t}\left(\begin{array}{c}
\cos2t\\
\cos2t+\sin2t\end{array}\right)$ and $x_{2}(t)=e^{t}\left(\begin{array}{c}
\sin2t\\
-\cos2t+\sin2t\end{array}\right)$.

Before we get to the general solution, let me describe how to draw
these two solutions.

$x_{1}^{\prime}(t)=e^{t}\left(\begin{array}{c}
\cos2t\\
\cos2t+\sin2t\end{array}\right)+e^{t}\left(\begin{array}{c}
-2\sin2t\\
-2\sin2t+2\cos2t\end{array}\right)$

Then $x_{1}(0)=\left(\begin{array}{c}
1\\
1\end{array}\right)$ and $x_{1}^{\prime}(0)=\left(\begin{array}{c}
1\\
1\end{array}\right)+\left(\begin{array}{c}
0\\
2\end{array}\right)=\left(\begin{array}{c}
1\\
3\end{array}\right)$

Thus start at $\left(\begin{array}{c}
1\\
1\end{array}\right)$, and start spiraling outwards in the direction of $\left(\begin{array}{c}
1\\
3\end{array}\right)$.

Note 1: Spiral outwards because $\Re\{\lambda\}=1>0$. Where $\Re\{\lambda\}$
means the real part of $\lambda$.

Similarly, $x_{2}^{\prime}(t)=x_{2}(t)+e^{t}\left(\begin{array}{c}
2\cos2t\\
2\sin2t+2\cos2t\end{array}\right)$

$x_{2}(0)=\left(\begin{array}{c}
0\\
-1\end{array}\right)$

$x_{2}^{\prime}(0)=\left(\begin{array}{c}
0\\
-1\end{array}\right)+\left(\begin{array}{c}
2\\
2\end{array}\right)=\left(\begin{array}{c}
2\\
1\end{array}\right)$