• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q6E

Expert-verified
Found in: Page 411

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the singular values of ${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$. Find a unit vector$\stackrel{\mathbf{\to }}{\mathbf{v}}$such that$||A\stackrel{\to }{{v}_{1}}||{\mathbf{=}}{{\mathbf{\sigma }}}_{{\mathbf{1}}}$. Sketch the image of the unit circle.

The singular values of A are ${\sigma }_{1}=5$ and ${\sigma }_{2}=0$

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

See the step by step solution

## Step 1: Determine eigen vector corresponding to the eigen value

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

$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{det}\left({\mathrm{xI}}_{2}-{\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right)\phantom{\rule{0ex}{0ex}}=\left(\mathrm{x}-5\right)\left(\mathrm{x}-20\right)-100\phantom{\rule{0ex}{0ex}}={\mathrm{x}}^{2}-25\mathrm{x}+100-100\phantom{\rule{0ex}{0ex}}={\mathrm{x}}^{2}-25\mathrm{x}$

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

Now,${\mathrm{x}}^{2}-25\mathrm{x}=\mathrm{x}\left(\mathrm{x}-25\right)$ . 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 ${\lambda }_{1}=25$ and ${\mathrm{\lambda }}_{2}=0$. Thus the singular values of A are ${\sigma }_{1}=\sqrt{25}=5$and ${\sigma }_{2}=\sqrt{0}=0$.

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

$\begin{array}{l}20{x}_{1}-10{x}_{2}=0\\ -10{x}_{1}+5{x}_{2}=0\end{array}$

This implies that ${x}_{2}=2{x}_{1}$.

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.

## Step 2: Substitute the equations

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

$\begin{array}{l}5{y}_{1}+10{y}_{2}=0\\ 10{y}_{1}+2{y}_{2}=0\end{array}$

This implies that ${y}_{1}=-2{y}_{2}$.

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 $||A{v}_{1}||=\sqrt{5+20}=\sqrt{25}=5={\sigma }_{1}$.

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

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: