Chapter 6 Exercise A

2. Solution: It does not satisfy definiteness. For the function takes $(0,1,0)$, $(0,1,0)$ to $0$, but $(0,1,0)\ne 0$.

4. Solution: (a) Note that $V$ is a real inner product space, we have $\langle u,v\rangle=\langle v,u\rangle$. Hence \begin{align*} \langle u+v,u-v\rangle&=\langle u,u\rangle-\langle u,v\rangle+\langle v,u\rangle-\langle v,v\rangle\\ &=\langle u,u\rangle-\langle v,v\rangle=\|u\|^2-\|v\|^2. \end{align*} (b) By (a).

(c) See the picture in Page 174 and note $\|u\|=\|v\|$ for a rhombus, then use (b).

5. Solution: Suppose $V$ is finite-dimensional here (I am not sure whether it is true for infinite-dimensional case). Hence we just need to show $T-\sqrt{2}I$ is injective. Suppose $u\in\m{null}(T-\sqrt{2}I)$, then \[Tu=\sqrt{2}u\Longrightarrow \|Tu\|=\sqrt{2}\|u\|.\]As $\|Tv\|\le \|v\|$ for every $v\in V$, it follows that $\|u\|=0$, hence $u=0$. That implies $T-\sqrt{2}I$ is injective.

6. Solution: If $\langle u,v\rangle =0$, then \[\|u+av\|^2=\|u\|^2+\|av\|^2\ge \|u\|^2\]by 6.13.

If $\|u\|\le \|u+av\|$ for all $a\in\mb F$, this implies \[ \|u+av\|^2-\|u\|=|a|^2\|v\|+a\langle v,u\rangle +\bar a\langle u,v\rangle\ge 0. \]If $v=0$, then $\langle u,v\rangle=0$. If $v\ne 0$, plug $a=-\langle u,v\rangle/\|v\|^2$ into the previous equation, we obtain \[ -\frac{|\langle u,v\rangle|^2}{\|v\|^2}\ge 0. \]Hence $\langle u,v\rangle=0$.

7. Solution: If $\|av+bu\|=\|au+bv\|$ for all $a,b\in\mb R$, by setting $a=1$ and $b=0$, we have $\|u\|=\|v\|$.

Conversely, suppose $\|u\|=\|v\|$. For all $a,b\in\mb R$, we have\begin{align*}\|av+bu\|^2=&\,\langle av+bu,av+bu\rangle\\ =&\, a^2\|u\|^2+ab(\langle u,v\rangle+\langle v,u\rangle)+b^2\|v\|^2\end{align*}and\begin{align*}\|au+bv\|^2=&\,\langle au+bv,au+bv\rangle\\ =&\, a^2\|v\|^2+ab(\langle u,v\rangle+\langle v,u\rangle)+b^2\|u\|^2.\end{align*}Hence if $\|v\|=\|u\|$, we have $$a^2\|u\|^2+b^2\|v\|^2=a^2\|v\|^2+b^2\|u\|^2.$$Therefore $\|av+bu\|^2=\|au+bv\|^2$, i.e. $\|av+bu\|=\|au+bv\|$.

8. Solution: Consider $\|u-v\|^2$, we have \begin{align*} \|u-v\|^2=&\langle u-v,u-v\rangle=\langle u,u\rangle-\langle u,v\rangle-\langle v,u\rangle+\langle v,v\rangle\\ =&\|u\|^2-\langle u,v\rangle-\overline{\langle u,v\rangle}+\|v\|^2=0, \end{align*} hence $u-v=0$ by definiteness. That is $u=v$.

9. Solution: By 6.15, we have $|\langle u,v\rangle |\leqslant \|u\|\|v\|$. Since $\|u\|\leqslant $ and $\|v\|\leqslant$, we also have\[0\leqslant 1-\|u\|\|v\|\leqslant 1-|\langle u,v\rangle |.\]To show $\sqrt{1-\|u\|^2}\sqrt{1-\|v\|^2}\leqslant 1-|\langle u,v\rangle |$, it suffices to show that\[\sqrt{1-\|u\|^2}\sqrt{1-\|v\|^2}\leqslant 1-\|u\|\|v\|.\] Since $0\leqslant 1-\|u\|\|v\|$, by squaring both sides, we only need to show\[(1-\|u\|^2)(1-\|v\|^2)\leqslant (1-\|u\|\|v\|)^2,\]which amounts to show\[(\|u\|-\|v\|)^2\geqslant 0.\]This completes the proof.

10. Solution: Let $v=(x,y)$ and $u=z(1,3)$, where $x,y,z\in \R$. Note that $v$ is orthogonal to $(1,3)$, we have \[ (x,y)\cdot (1,3)=x+3y=0. \]It follows that $v=x(-3,1)$. Since $(1,2)=u+v$, we obtain \[ x(-3,1)+z(1,3)=(z-3x,x+3z)=(1,2). \]We can solve this equation and get $x=-1/10$ and $z=7/10$. Hence $u=(7/10,21/10)$ and $v=(3/10,-1/10)$.

11. Solution: Consider Example 6.17 (a), we have \begin{align*} &(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\\ \geqslant &\left(\sqrt{a\times\frac{1}{a}}+\sqrt{b\times\frac{1}{b}}+\sqrt{c\times\frac{1}{c}}+\sqrt{d\times\frac{1}{d}}\right)^2\\ =& 4^2=16. \end{align*}

12. Solution: In Example 6.17 a), let $y_i=1$.

15. Solution: Consider Example 6.17 (a). Let $x_j=\sqrt{j|a_j|}$ and $y_j=\sqrt{\frac{|b_j|}{j}}$ and note that
\[|a_1b_1+\cdots+a_nb_n|\leqslant \sum_{j=1}^n |a_jb_j|.\]

16. Solution: Note that (…..After I finished this, I found that is exactly 6.22…) \begin{align*} &\|u+v\|^2+\|u-v\|^2 \\ =&\langle u+v,u+v\rangle +\langle u-v,u-v\rangle\\ =&\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle+\langle u,u\rangle-\langle u,v\rangle-\langle v,u\rangle+\langle v,v\rangle\\ =&2\langle u,u\rangle+2\langle v,v\rangle=2\|u\|^2+2\|v\|^2, \end{align*} it follows that \[2\times 3^2+2\|v\|^2=4^2+6^2.\]Hence $\|v\|=\sqrt{17}$.

17. Solution: By 6.22, if there is such an inner product on $\R^2$, then we must have \[ \|\alpha-\beta\|^2+\|\alpha+\beta\|^2=2(\|\alpha\|^2+\|\beta\|^2). \]Let $\alpha=(1,0)$ and $\beta=(0,1)$, we will get a counterexample.

19. Solution: See it here Exercise 1 or See Linear Algebra Done Right Solution Manual Chapter 6 Problem 6.

20. Solution: See it here Exercise 1 or See Linear Algebra Done Right Solution Manual Chapter 6 Problem 7.

21. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 8.

22. Solution: It follows directly from Problem 12.

24. Solution: Positivity: $\langle u,u\rangle_1=\langle Su,Su\rangle\ge 0$ for all $u\in V$.

Definiteness: $0=\langle u,u\rangle_1=\langle Su,Su\rangle$, hence $Su=0$. As $S$ is injective, it follows that $u=0$.

Additivity in first slot: \begin{align*} \langle u+v,w\rangle_1=&\langle S(u+v),Sw\rangle=\langle Su+Sv,Sw\rangle\\ =&\langle Su,Sw\rangle+\langle Sv,Sw\rangle=\langle u,w\rangle_1+\langle v,w\rangle_1. \end{align*} Homogeneity in first slot: \begin{align*} \langle \lambda u,w\rangle_1=&\langle S(\lambda u),Sw\rangle=\langle \lambda Su,Sw\rangle\\ =&\lambda\langle Su,Sw\rangle=\lambda\langle u,w\rangle_1. \end{align*} Conjugate symmetry: $\langle u,v\rangle_1=\langle Su,Sv\rangle=\overline{\langle Sv,Su\rangle}=\overline{\langle v,u\rangle_1}$.

25. Solution: Note that $S\in\ca L(V)$ is not injective, there exists a nonzero $u\in V$ such that $Su=0$. Now we have $\langle u,u\rangle_1=\langle Su,Su\rangle=0$, this implies that $\langle u,v\rangle_1$ do not satisfy definiteness.

27. Solution: Let $a=(w-u)/2$ and $b=(w-v)/2$, by 6.22, we have \[ \|a-b\|^2+\|a+b\|^2=2\|a\|^2+2\|b\|^2. \]Plug $a=(w-u)/2$ and $b=(w-v)/2$ into the expression above, we get\[\left\|w-\frac{1}{2}(u+v)\right\|^2=\frac{\|w-u\|^2+\|w-v\|^2}{2}-\frac{\|u-v\|^2}{4}.\]

28. Solution: Suppose there are two such vectors in $C$. Denote them by $\xi$ and $\mu$ ($\xi\ne\mu$), then we have \[\|w-\xi\|\le \|w-\mu\|\text{ and }\|w-\mu\|\le \|w-\xi\|\]by the choice of $\xi$ and $\mu$. Hence $\|w-\xi\|=\|w-\mu\|$. By the previous exercise, we have \[\left\|w-\frac{1}{2}(\xi+\mu)\right\|^2=\frac{\|w-\xi\|^2+\|w-\mu\|^2}{2}-\frac{\|\xi-\mu\|^2}{4}<\|w-\xi\|^2.\]This contradicts with the choice of $\xi$. Hence there is at most one $u\in C$ such that \[\|w-u\|\le \|w-v\|\quad \text{for all } v\in C.\]

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
This entry was posted in Chapter 6 and tagged .