91Ó°ÊÓ

$\mathrm{q}\left({\mathrm{x}}_1.{\mathrm{x}}_2\right)={\mathrm{x}}_1{\mathrm{x}}_2$

$\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{X}}_2\right)={\mathrm{x}}_1{\mathrm{x}}_2=1$

$=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\left[\begin{array}{c}\frac{1}{2}{x}_2\\ \frac{1}{2}{x}_1\end{array}\right]$

Split the term ${\mathrm{x}}_1{\mathrm{x}}_2$equally between the two components. Therefore

$\mathrm{q}\left(\stackrel{→}{\mathrm{x}}\right)=\stackrel{→}{\mathrm{x}}×\mathrm{A}\stackrel{→}{\mathrm{x}},\mathrm{where}\mathrm{A}=\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{2}& 0\end{array}\right]$

To determine the eigen values of the matrix A

$\det \left(\mathrm{A}-\mathrm{λ\pm ô}\right)=0\left|\begin{array}{cc}0-\mathrm{λ}& \frac{1}{2}\\ \frac{1}{2}& 0-\mathrm{λ}\end{array}\right|=0$

$\left(0-\mathrm{λ}\right)\left(0-\mathrm{λ}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=0$

${\mathrm{λ}}^2-\frac{1}{4}=0$

$\begin{array}{l}\mathrm{λ}=\sqrt{\frac{1}{4}}\\ =Â\pm \frac{1}{2}\end{array}$

The eigen values are ${\mathrm{λ}}_1=\frac{1}{2}$and ${\mathrm{λ}}_2=-\frac{1}{2}$

To find the eigen vectors are ${\mathrm{λ}}_1=\frac{1}{2}$

$$\begin{gathered} \left(\mathrm{A}-\frac{1}{2}\mathrm{l}\right)\tilde{\mathrm{x}}=0\left[\begin{array}{cc}0-\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 0-\frac{1}{2}\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0 \\ \left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0 \end{gathered}$$

Apply Row operation ${\mathrm{R}}_2→{\mathrm{R}}_2+{\mathrm{R}}_1$

$\left[\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0$

The first row implies ${\mathrm{R}}_1→2{\mathrm{R}}_1$

$\left[\begin{array}{cc}-1& 1\\ 0& 0\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0$

${\mathrm{x}}_1={\mathrm{x}}_2,{\mathrm{x}}_2=1,{\mathrm{x}}_1=1,{\mathrm{λ}}_1=\frac{1}{2}$

$\stackrel{→}{{\mathrm{u}}_1}=\left[\begin{array}{c}1\\ 1\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{→}{\mathrm{v}}}_1=\frac{1}{||{\stackrel{→}{\mathrm{u}}}_1||}{\stackrel{→}{\mathrm{u}}}_1=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]$

When ${\mathrm{λ}}_2=-\frac{1}{2}$

$$\begin{gathered} \left(\mathrm{A}+\frac{1}{2}\mathrm{l}\right)\tilde{\mathrm{x}}=0\left[\begin{array}{cc}0+\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 0+\frac{1}{2}\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0 \\ \left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0 \end{gathered}$$

Apply row operation ${\mathrm{R}}_2→{\mathrm{R}}_2-{\mathrm{R}}_1$

$\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=0$

${\mathrm{x}}_1={\mathrm{x}}_2,{\mathrm{x}}_2=-1,{\mathrm{x}}_1=1,{\mathrm{λ}}_1=-\frac{1}{2}$

${\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}1\\ -1\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{→}{\mathrm{v}}}_2=\frac{1}{||{\stackrel{→}{\mathrm{u}}}_1||}{\stackrel{→}{\mathrm{u}}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]$

$\frac{1}{2}{\mathrm{c}}_1^2-\frac{1}{2}{\mathrm{c}}_2^2=1$

$\stackrel{→}{{\mathrm{v}}_1}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right],\stackrel{→}{{\mathrm{v}}_2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]$

The graph of the coordinate axes is