91Ó°ÊÓ

Given that

$A=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$

Now the singular values of A are found by first computing the eigenvalues of the square matrix

${\mathrm{A}}^{\mathrm{t}}\mathrm{A}=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}5& 10\\ 10& 20\end{array}\right]$

Then

$\left({\mathrm{xI}}_2-{\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right)=\left[\begin{array}{cc}\mathrm{x}-5& -10\\ -10& \mathrm{x}-20\end{array}\right]$

Let characteristic polynomial of ${\mathrm{A}}^{\mathrm{t}}\mathrm{A}$be $\mathrm{p}\left(\mathrm{x}\right)$. Then

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\det ({\mathrm{xI}}_2-{\mathrm{A}}^{\mathrm{t}}\mathrm{A}) \\ =(\mathrm{x}-5)\left(\mathrm{x}-20\right)-100 \\ ={\mathrm{x}}^2-25\mathrm{x}+100-100 \\ ={\mathrm{x}}^2-25\mathrm{x} \end{gathered}$$

Thus the characteristic polynomial is $x^2-25x$.

Now,${\mathrm{x}}^2-25\mathrm{x}=\mathrm{x}(\mathrm{x}-25)$. This implies that the roots of $\mathrm{p}\left(\mathrm{x}\right)$are 25,0 . Hence the eigenvalues of ${\mathrm{A}}^{\mathrm{t}}\mathrm{A}$are ${\mathrm{λ}}_1=25$and ${\mathrm{λ}}_2=0$. Thus the singular values of A are ${\mathrm{σ}}_1=\sqrt{25}=5$and ${\mathrm{σ}}_2=\sqrt{0}=0$.

Now let

$X_1=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

be an eigenvector corresponding to the eigenvalue 25. Therefor $\left({\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right){\mathrm{X}}_1=25{\mathrm{X}}_1$which means $\left(25{\mathrm{I}}_2-{\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right){\mathrm{X}}_1=0$. From this equation we get that

$\left[\begin{array}{cc}20& -10\\ -10& 5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Therefore

$\begin{array}{l}20x_1-10x_2=0\\ -10x_1+5x_2=0\end{array}$

This implies that$x_2=2x_1$.

Therefore

${\mathrm{x}}_2=2{\mathrm{x}}_1\left[\begin{array}{c}\mathrm{r}\\ 2\mathrm{r}\end{array}\right]$

is an eigenvector corresponding to the eigenvalue 25 for any non-zero real number r. By putting $r=1$, we conclude that

$\left[\begin{array}{c}1\\ 2\end{array}\right]$

is an eigenvector corresponding to the eigenvalue 25.

Now let

$X_2=\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]$

be an eigenvector corresponding to the eigenvalue 0. Therefore $\left({\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right){\mathrm{X}}_1=\mathrm{O}$. From this equation we get that

$\left[\begin{array}{cc}5& 10\\ 10& 20\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Therefore

$\begin{array}{l}5y_1+10y_2=0\\ 10y_1+2y_2=0\end{array}$

This implies that $y_1=-2y_2$.

Therefore

$\left[\begin{array}{c}-2r\\ r\end{array}\right]$

is an eigenvector corresponding to the eigenvalue 0 for any non-zero real number. By putting r=1, we conclude that

$\left[\begin{array}{c}-2\\ 1\end{array}\right]$

is an eigenvector corresponding to the eigenvalue 0. an eigenvector corresponding to the

Let

$v_1=\frac{1}{\sqrt{5}}{X}_1=\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{array}\right]$

and

let,

$v_2=\frac{1}{\sqrt{5}}{X}_2=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]=\left[\begin{array}{c}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{array}\right]$

Then

$A{v}_1=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{array}\right]=\left[\begin{array}{c}\sqrt{5}\\ 2\sqrt{5}\end{array}\right]$

Hence $||A{v}_1||=\sqrt{5+20}=\sqrt{25}=5={\mathrm{σ}}_1$.

$v_1=\left[\begin{array}{c}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{array}\right]$

is the required vector.

Also,

$L\left(v_2\right)=A{v}_2\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\left[\begin{array}{c}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Now by using Theorem 8.3.2 we get that, the image of the unit circle is an ellipse whose semi-major and semi-minor axes have lengths 5 and 0 respectively. Hence the image of the unit circle is a straight line of the following form: