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\begin{document}


{\bf Question}

Calculate the eigenvalues and the corresponding eigenvectors for
each of the following matrices:

$${\rm (i)}\left( \begin{array}{rr}
     1 &amp; {-1} \\
     0 &amp; 2
 \end{array}  \right);\ \
 {\rm (ii)} \left( \begin{array}{rr}
     3 &amp; 5 \\
     {-1} &amp; 2
 \end{array}  \right);\ \
 {\rm (iii)} \left( \begin{array}{rr}
     4 &amp; 1 \\
     {-1} &amp; {-2}
 \end{array}  \right);$$

 $${\rm (iv)} \left( \begin{array}{rr}
     {\cos (\theta )} &amp; {\sin (\theta )} \\
     {-\sin (\theta )} &amp; {\cos (\theta )}
 \end{array}  \right);\ \ \
 {\rm (v)} \left( \begin{array}{rr}
     {\cos (2\theta )} &amp; {\sin (2\theta )} \\
     {\sin (2\theta )} &amp; {-\cos (2\theta )}
 \end{array}  \right).$$



{\bf Answer}

\begin{description}
\item[(i)]
$\left|\begin{array}{ll}{1- \lambda} &amp; -1\\ 0 &amp;
{2-\lambda}\end{array} \right|=(1-\lambda)(2-\lambda)=0
\Rightarrow \lambda = 1,2.$

\begin{description}
\item[\underline {$\lambda=1$}]
Solve $\left(\begin{array}{rr} 0 &amp; {-1}\\ 0 &amp; 1
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$ or $y=0$

\ \ \ Let $x=\alpha,\ y=0$ so a suitable eigenvector
$\left(\begin{array}{c} \alpha \\ 0
\end{array}\right)$

\item[\underline {$\lambda=2$}]
Solve $\left(\begin{array}{rr} -1 &amp; -1\\ 0 &amp; 0
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$

\ \ \ or $-x-y=0 \Rightarrow y=-x$

\ \ \ Let $x=\beta,\ y=-\beta$ so a suitable eigenvector
$\left(\begin{array}{c} \beta \\ -\beta
\end{array}\right)$
\end{description}

\item[(ii)]
$\left|\begin{array}{cc} 3-\lambda &amp; 5\\ -1 &amp; 2-\lambda
\end{array}\right|=(3-\lambda)(2-\lambda)+5=\lambda^2-5\lambda+11=0$
$$\lambda=\frac{5\pm \sqrt{25-44}}{2}=\frac{5\pm \sqrt{19}i}{2}$$

No real roots, so no real eigenvectors.

\item[(iii)]
$\left|\begin{array}{cc} 4-\lambda &amp; 1\\ -1 &amp; -2-\lambda
\end{array}\right|=(4-\lambda)(-2-\lambda)+1=\lambda^2-2\lambda-7=0$
$$\lambda=\frac{2\pm \sqrt{32}}{2}=\frac{2\pm 4\sqrt{2}}{2}=1\pm
2\sqrt{2}$$

\underline {$\lambda = 1 + 2\sqrt{2}$}

Solve $\left(\begin{array}{cc} 3-2\sqrt{2} &amp; 1\\ -1 &amp; -3-2\sqrt{2}
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$

So $\left.\begin{array} {rcl} {(3-2\sqrt{2})x+y} &amp; = &amp; 0\\
{-x-(3+2\sqrt{2})y} &amp; = &amp; 0 \end{array}\right\}$

Let $x=\alpha$, so $y=-\alpha(3-2\sqrt{2})$

Note from second equation:

$\displaystyle
y=\frac{-\alpha}{3+2\sqrt{2}}=\frac{-\alpha(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}
=\frac{-\alpha(3-2\sqrt{2})}{1}$

Suitable eigenvector: $\left(\begin{array}{cc} \alpha \\
\alpha(2\sqrt{2}-3)\end{array}\right)$.

\underline {$\lambda = 1 - 2\sqrt{2}$}

Solve $\left(\begin{array}{cc} 3+2\sqrt{2} &amp; 1\\ -1 &amp; -3+2\sqrt{2}
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$

So $\left.\begin{array} {rcl} {(3+2\sqrt{2})x+y} &amp; = &amp; 0\\
{-x+(-3+2\sqrt{2})y} &amp; = &amp; 0 \end{array}\right\}$

Let $x=\beta$, so $y=-\beta(3+2\sqrt{2})$

Suitable eigenvector: $\left(\begin{array}{cc} \beta \\
-\beta(3+2\sqrt{2})\end{array}\right)$.

\item[(iv)]
$\left|\begin{array}{cc} \cos \theta-\lambda &amp; \sin \theta\\ -\sin
\theta &amp; \cos \theta-\lambda
\end{array}\right|=(\cos \theta-\lambda)^2+\sin^2 \theta=$

$\lambda^2-2\lambda \cos \theta+\cos^2 \theta+\sin^2
\theta=\lambda^2 - 2\lambda \cos \theta+1=0$

$$\lambda=\frac{2\cos \theta\pm \sqrt{4\cos^2 \theta-4}}{2}=\cos
\theta \pm i \sin \theta.$$

If $\sin \theta \ne 0$, then no real solutions, so no real
eigenvectors.

If $\sin \theta = 0$, then repeated eigenvalue $\lambda=\cos
\theta$ with eigenvector equation

$$\left(\begin{array}{cc} 0 &amp; 0\\ 0 &amp; 0 \end{array} \right)
\left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$$

i.e. $\left.\begin{array} {rcl} {\rm {any}\ x} &amp; \Rightarrow &amp;
{x=\alpha}\\ {\rm {any}\ y} &amp; \Rightarrow &amp; {y=\beta}
\end{array}\right\}$  eigenvector $\left(\begin{array}{c} \alpha \\
\beta \end{array} \right)$

(So any non-zero vector in the xy-plane is a suitable eigenvector)

\item[(v)]
$\left|\begin{array}{cc} \cos 2\theta-\lambda &amp; \sin 2\theta\\
-\sin 2\theta &amp; \cos 2\theta-\lambda
\end{array}\right|=
\lambda^2-\cos^2 2\theta-\sin^2 2\theta=\lambda^2 - 1=0$

so $\lambda = \pm 1$.

\underline {Assume $\sin 2\theta \ne 0$}

\begin{description}
\item[\underline{$\lambda =1$}]
Solve $\left(\begin{array}{cc} \cos 2 \theta - 1 &amp; \sin 2\theta\\
\sin 2 \theta &amp; -\cos 2\theta-1
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$

$\left.\begin{array} {rcl} (\cos 2\theta -1)x+(\sin 2\theta)y  &amp; =
&amp; 0\\ (\sin 2\theta)x-(\cos 2\theta + 1)y &amp; = &amp; 0
\end{array}\right\}$

Let $x=\alpha$ so $\displaystyle y=\frac{(1-\cos
2\theta)\alpha}{\sin 2\theta}$ ($\sin 2\theta \ne 0$)

Suitable eigenvector: $\displaystyle \left(\begin{array}{c} \alpha
\\ \frac{(1-\cos 2\theta)\alpha}{\sin 2\theta} \end{array}
\right)$

Note from second equation:

\begin{eqnarray*}y=\frac{\alpha (\sin 2\theta)} {\cos 2\theta +
1}&amp; = &amp; \frac{\alpha(\sin 2 \theta)}{\cos 2\theta + 1}
\left(\frac{1-\cos 2\theta}{1-\cos 2\theta}\right)\\ &amp; = &amp;
\frac{\alpha(\sin 2 \theta)(1-\cos 2\theta)}{1-\cos^2 2\theta}\\ &amp;
= &amp; \frac{\alpha(\sin 2 \theta)(1-\cos 2\theta)}{\sin^2 2\theta}\\
&amp; = &amp; \frac{(1-2\cos 2 \theta)\alpha}{\sin 2\theta}
\end{eqnarray*}

\item[\underline{$\lambda =-1$}]
Solve $\left(\begin{array}{cc} \cos 2 \theta + 1 &amp; \sin 2\theta\\
\sin 2 \theta &amp; \cos 2\theta+1
\end{array}\right) \left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)$

$\left.\begin{array} {rcl} (\cos 2\theta +1)x+(\sin 2\theta)y  &amp; =
&amp; 0\\ (\sin 2\theta)x+(1-\cos 2\theta)y &amp; = &amp; 0
\end{array}\right\}$

Let $x=\beta$ so $\displaystyle y=\frac{-\beta(1+\cos
2\theta)}{\sin 2\theta}$ ($\sin 2\theta \ne 0$)

Suitable eigenvector: $\displaystyle \left(\begin{array}{c} \beta
\\ \frac{-\beta(1+\cos 2\theta)}{\sin 2\theta} \end{array}
\right)$

\end{description}

\begin{description}
\item[\underline{\rm {If} $\sin 2\theta =0$}]
$$\rm{The\ matrix\ becomes}\ \left(\begin{array}{cc} \cos 2 \theta
&amp; 0\\ 0 &amp; -\cos 2 \theta \end{array} \right)\ \rm{with\
eigenvalues}$$

$$(\cos 2\theta - \lambda)(-\cos 2\theta -
\lambda)=\lambda^2-\cos^2 2\theta=0 \Rightarrow \lambda=\pm \cos
2\theta.$$

\item[\underline{$\lambda = \cos 2\theta$}]

$$\left(\begin{array}{cc} 0 &amp; 0\\0 &amp; -2 \cos 2\theta \end{array}
\right)\left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)\ \rm{so}\ (-2 \cos 2\theta)y=0\ \rm{and}\ y=0$$

If $x=\alpha$ then suitable eigenvector: $\left(\begin{array}{c}
\alpha \\ 0 \end{array} \right)$

\item[\underline{$\lambda = -\cos 2\theta$}]

$$\left(\begin{array}{cc} 2\cos 2\theta &amp; 0\\0 &amp; 0
\end{array} \right)\left(\begin{array}{c} x\\ y
\end{array}\right)=\left(\begin{array}{c} 0\\ 0
\end{array}\right)\ \rm{so}\ (2 \cos 2\theta)x=0\ \rm{and}\ x=0$$

If $y=\beta$ then suitable eigenvector: $\left(\begin{array}{c} 0
\\ \beta \end{array} \right)$

\end{description}
\end{description}



\end{document}
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