Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the singular values of $A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}]$ . Find a unit vector $\vec{v}$ such that $|| A \vec{v_{1}} || = σ_{1}$ . Sketch the image of the unit circle.

The singular values of A are $σ_{1} = 5$ and $σ_{2} = 0$

$v_{1} = [\begin{matrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}]$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine eigen vector corresponding to the eigen value

Given that

$A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}]$

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

$A^{t} A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] = [\begin{matrix} 5 & 10 \\ 10 & 20 \end{matrix}]$

Then

$({xI}_{2} - A^{t} A) = [\begin{matrix} x - 5 & - 10 \\ - 10 & x - 20 \end{matrix}]$

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

$p (x) = \det ({xI}_{2} - A^{t} A) = (x - 5) (x - 20) - 100 = x^{2} - 25 x + 100 - 100 = x^{2} - 25 x$

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

Now, $x^{2} - 25 x = x (x - 25)$ . This implies that the roots of $p (x)$ are 25,0 . Hence the eigenvalues of $A^{t} A$ are $λ_{1} = 25$ and $λ_{2} = 0$ . Thus the singular values of A are $σ_{1} = \sqrt{25} = 5$ and $σ_{2} = \sqrt{0} = 0$ .

Now let

$X_{1} = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]$

be an eigenvector corresponding to the eigenvalue 25. Therefor $(A^{t} A) X_{1} = 25 X_{1}$ which means $(25 I_{2} - A^{t} A) X_{1} = 0$ . From this equation we get that

$[\begin{matrix} 20 & - 10 \\ - 10 & 5 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

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

$x_{2} = 2 x_{1} [\begin{matrix} r \\ 2 r \end{matrix}]$

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

$[\begin{matrix} 1 \\ 2 \end{matrix}]$

is an eigenvector corresponding to the eigenvalue 25.

Step 2: Substitute the equations

Now let

$X_{2} = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$

be an eigenvector corresponding to the eigenvalue 0. Therefore $(A^{t} A) X_{1} = O$ . From this equation we get that

$[\begin{matrix} 5 & 10 \\ 10 & 20 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

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

$[\begin{matrix} - 2 r \\ r \end{matrix}]$

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

$[\begin{matrix} - 2 \\ 1 \end{matrix}]$

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

Let

$v_{1} = \frac{1}{\sqrt{5}} X_{1} = \frac{1}{\sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}] = [\begin{matrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}]$

and

let,

$v_{2} = \frac{1}{\sqrt{5}} X_{2} = \frac{1}{\sqrt{5}} [\begin{matrix} - 2 \\ 1 \end{matrix}] = [\begin{matrix} - \frac{2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \end{matrix}]$

Then

$A v_{1} = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}] = [\begin{matrix} \sqrt{5} \\ 2 \sqrt{5} \end{matrix}]$

Hence $||A v_{1}|| = \sqrt{5 + 20} = \sqrt{25} = 5 = σ_{1}$ .

$v_{1} = [\begin{matrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}]$

is the required vector.

Also,

$L (v_{2}) = A v_{2} [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] [\begin{matrix} - \frac{2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

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:

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Linear Algebra With Applications

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Short Answer

Find the singular values of $A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}]$ . Find a unit vector $\vec{v}$ such that $|| A \vec{v_{1}} || = σ_{1}$ . Sketch the image of the unit circle.

Step by Step Solution

Step 1: Determine eigen vector corresponding to the eigen value

Step 2: Substitute the equations

Step 3: Mapping the graph

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Step by Step Solution

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Step 2: Substitute the equations

Step 3: Mapping the graph

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Find the singular values of $A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}]$ . Find a unit vector $\vec{v}$ such that $|| A \vec{v_{1}} || = σ_{1}$ . Sketch the image of the unit circle.