Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

51. If A is a symmetric $2 \times 2$ matrix with eigenvalues 1 and 2, then the angle between $\vec{x}$ and $A \vec{x}$ must be less than $π / 6$ , for all nonzero vectors $\vec{x}$ in $R^{2}$ .

The given statement is TRUE.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Check whether the given statement is TRUE or FALSE

Let ${\dot{v}}_{1}, {\dot{v}}_{2}$ be unit orthogonal eigenvectors with eigenvalues 1 and 2 respectively.

So, write localid="1659679849030" $\dot{x} = c_{1} {\dot{v}}_{1} + c_{2} {\dot{v}}_{2} for some c_{1}, c_{2} . Also ||\dot{x}|| = \sqrt{c_{1}^{2} + c_{2}^{2}}$

$\Rightarrow A \dot{x} = c_{1} A {\dot{v}}_{1} + c_{2} A {\dot{v}}_{2} \Rightarrow A \dot{x} = c_{1} {\dot{v}}_{1} + 2 c_{1} {\dot{v}}_{2} \Rightarrow ||\dot{x}|| = \sqrt{c_{1}^{2} + 4 c_{2}^{2}} \Rightarrow \dot{x} . A \dot{v} = c_{1}^{2} + 2 c_{2}^{2}$

Let is the angle between $\dot{x}$ and $A \dot{x}$ . Then,

$\cos θ = \frac{\dot{x} . A \dot{a}}{||\dot{x}|| ||A \dot{x}||} = \frac{c_{1}^{2} + 2 c_{2}^{2}}{\sqrt{c_{1}^{2} + c_{2}^{2}} \sqrt{c_{1}^{2} + 4_{2}^{2}}} = \frac{\sqrt{(c_{1}^{2} + 2 c_{2}^{2})}}{\sqrt{c_{1}^{4} + 5 c_{1}^{2} c_{2}^{2} + 4 c_{2}^{4}}} = {(\frac{c_{1}^{4} + 4 c_{1}^{2} c_{2}^{2} + 4 c_{2}^{4}}{c_{1}^{4} + 4 c_{1}^{2} c_{2}^{2} + 4 c_{2}^{4} + c_{1}^{2} c_{2}^{2}})}^{1 / 2}$

$= (\frac{1}{1 + \frac{c_{1}^{2} c_{2}^{2}}{c_{1}^{4} + 4 c_{1}^{2} c_{2}^{2} + 4 c_{2}^{4}}}) = (\frac{1}{1 + \frac{c_{1}^{2} c_{2}^{2}}{c_{1}^{4} + 4 c_{1}^{2} c_{2}^{2} + 4 c_{2}^{4}}}) = {(\frac{1}{1 + \frac{c_{1}^{2} c_{2}^{2}}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}}})}^{1 / 2} = {(\frac{1}{1 + \frac{c_{1} c_{2}}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}}})}^{1 / 2}$

Here, $\cos θ$ takes minimum value when localid="1659678852211" $\frac{c_{1} c_{2}}{c_{1}^{2} + c_{2}^{2}}$ takes maximum value.

To find the maximum value of $\frac{c_{1} c_{2}}{c_{1}^{2} + c_{2}^{2}}$ , differentiate it with respect to $c_{1}$ and $c_{2}$ .

$\frac{\partial}{\partial c_{1}} (\frac{c_{1} c_{2}}{c_{1}^{2} + 2 c_{2}^{2}}) = \frac{(c_{1}^{2} + 2 c_{2}^{2}) c_{2} - c_{1} c_{2} (2 c_{2})}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}} = \frac{- c_{1}^{2} c_{2} + 2 c_{2}^{3}}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}} = \frac{c_{2} (- c_{1}^{2} + 2 c_{2}^{2})}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}} \frac{\partial}{\partial c_{1}} (\frac{c_{1} c_{2}}{c_{1}^{2} + 2 c_{2}^{2}}) = \frac{(c_{1}^{2} + 2 c_{2}^{2}) c_{2} - c_{1} c_{2} (4 c_{2})}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}} = \frac{c_{1}^{3} - 2 c_{1} c_{2}^{2}}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}} = \frac{c_{2} (c_{1}^{2} - 2 c_{2}^{2})}{{(c_{1}^{2} + 2 c_{2}^{2})}^{2}}$

In order to find the extreme value, equate both the partial derivatives with 0. Ignore the solution $c_{1}, c_{2} = 0$ since it corresponds to minimum. The other condition we obtain is $c_{1}^{2} = 2 c_{2}^{2}$ . Thus, the maximum is achieved when this condition is satisfied.

When $c_{1}^{2} = 2 c_{2}^{2}$ , we have

$\frac{c_{1}^{2} c_{2}^{2}}{(c_{1}^{2} + 2 c_{2}^{2})} = \frac{2 c_{2}^{4}}{{(4 c_{2}^{2})}^{2}} = \frac{2 c_{2}^{4}}{16 c_{2}^{4}} = \frac{1}{8}$

Hence, $cosθ = {(\frac{1}{1 + {(1 / 8)}^{2}})}^{1 / 2} = \sqrt{8 / 9}$

Step 2: Final Answer

This means that the minimum possible value of is $\cos θ is \sqrt{8 / 9} > \sqrt{3} / 2$ . Thus, in any case $\cos θ > \sqrt{3} / 2$ which implies that $θ < π / 6$ .

Thus, the given statement is TRUE.

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Short Answer

51. If A is a symmetric $2 \times 2$ matrix with eigenvalues 1 and 2, then the angle between $\vec{x}$ and $A \vec{x}$ must be less than $π / 6$ , for all nonzero vectors $\vec{x}$ in $R^{2}$ .

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Linear Algebra With Applications

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51. If A is a symmetric 2×2 matrix with eigenvalues 1 and 2, then the angle between x→ and Ax→ must be less than π/6, for all nonzero vectors x→ in R2.

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Step 1: Check whether the given statement is TRUE or FALSE

Step 2: Final Answer

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51. If A is a symmetric $2 \times 2$ matrix with eigenvalues 1 and 2, then the angle between $\vec{x}$ and $A \vec{x}$ must be less than $π / 6$ , for all nonzero vectors $\vec{x}$ in $R^{2}$ .