Suggested languages for you:

Americas

Europe

Q51E

Expert-verified
Found in: Page 414

### Linear Algebra With Applications

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

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

The given statement is TRUE.

See the step by step solution

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

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

So, write localid="1659679849030" $\stackrel{˙}{\mathrm{x}}={\mathrm{c}}_{1}{\stackrel{˙}{\mathrm{v}}}_{1}+{\mathrm{c}}_{2}{\stackrel{˙}{\mathrm{v}}}_{2}\mathrm{for}\mathrm{some}{\mathrm{c}}_{1},{\mathrm{c}}_{2}.\mathrm{Also}\left|\left|\stackrel{˙}{\mathrm{x}}\right|\right|=\sqrt{{\mathrm{c}}_{1}^{2}+{\mathrm{c}}_{2}^{2}}$

$⇒\mathrm{A}\stackrel{˙}{\mathrm{x}}={\mathrm{c}}_{1}\mathrm{A}{\stackrel{˙}{\mathrm{v}}}_{1}+{\mathrm{c}}_{2}\mathrm{A}{\stackrel{˙}{\mathrm{v}}}_{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{A}\stackrel{˙}{\mathrm{x}}={\mathrm{c}}_{1}{\stackrel{˙}{\mathrm{v}}}_{1}+2{\mathrm{c}}_{1}{\stackrel{˙}{\mathrm{v}}}_{2}\phantom{\rule{0ex}{0ex}}⇒\left|\left|\stackrel{˙}{\mathrm{x}}\right|\right|=\sqrt{{\mathrm{c}}_{1}^{2}+4{\mathrm{c}}_{2}^{2}}\phantom{\rule{0ex}{0ex}}⇒\stackrel{˙}{\mathrm{x}}.\mathrm{A}\stackrel{˙}{\mathrm{v}}={\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}$

Let is the angle between $\stackrel{˙}{\mathrm{x}}$and $\mathrm{A}\stackrel{˙}{\mathrm{x}}$. Then,

$\mathrm{cos}\theta =\frac{\stackrel{˙}{x}.A\stackrel{˙}{a}}{||\stackrel{˙}{x}||||A\stackrel{˙}{x}||}\phantom{\rule{0ex}{0ex}}=\frac{{c}_{1}^{2}+2{c}_{2}^{2}}{\sqrt{{c}_{1}^{2}+{c}_{2}^{2}}\sqrt{{c}_{1}^{2}+{4}_{2}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{\left({c}_{1}^{2}+2{c}_{2}^{2}\right)}}{\sqrt{{c}_{1}^{4}+5{c}_{1}^{2}{c}_{2}^{2}+4{c}_{2}^{4}}}\phantom{\rule{0ex}{0ex}}={\left(\frac{{c}_{1}^{4}+4{c}_{1}^{2}{c}_{2}^{2}+4{c}_{2}^{4}}{{c}_{1}^{4}+4{c}_{1}^{2}{c}_{2}^{2}+4{c}_{2}^{4}+{c}_{1}^{2}{c}_{2}^{2}}\right)}^{1/2}$

Here, $\mathrm{cos}\mathrm{\theta }$ takes minimum value when localid="1659678852211" $\frac{{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{{\mathrm{c}}_{1}^{2}+{\mathrm{c}}_{2}^{2}}$takes maximum value.

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

$\frac{\partial }{\partial {c}_{1}}\left(\frac{{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{{\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}}\right)=\frac{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right){c}_{2}-{c}_{1}{c}_{2}\left(2{c}_{2}\right)}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-{c}_{1}^{2}{c}_{2}+2{c}_{2}^{3}}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{c}_{2}\left(-{c}_{1}^{2}+2{c}_{2}^{2}\right)}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\partial }{\partial {c}_{1}}\left(\frac{{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{{\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}}\right)=\frac{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right){c}_{2}-{c}_{1}{c}_{2}\left(4{c}_{2}\right)}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{c}_{1}^{3}-2{c}_{1}{c}_{2}^{2}}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{c}_{2}\left({c}_{1}^{2}-2{c}_{2}^{2}\right)}{{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}^{2}}$

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

When ${\mathrm{c}}_{1}^{2}=2{\mathrm{c}}_{2}^{2}$, we have

$\frac{{\mathrm{c}}_{1}^{2}{\mathrm{c}}_{2}^{2}}{\left({\mathrm{c}}_{1}^{2}+2{\mathrm{c}}_{2}^{2}\right)}=\frac{2{\mathrm{c}}_{2}^{4}}{{\left(4{\mathrm{c}}_{2}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2{\mathrm{c}}_{2}^{4}}{16{\mathrm{c}}_{2}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}$

Hence, $\mathrm{cos\theta }={\left(\frac{1}{1+{\left(1/8\right)}^{2}}\right)}^{1/2}=\sqrt{8/9}$

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