Chapter 6 Exercise A


1. Solution: It does not satisfy homogeneity on the first slot when $\lambda$ is negative.


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


3. Solution: Let us show positivity by contradiction. Suppose there exists $w\in V$ such that $\langle w,w\rangle <0$. Let $v\in V$ be such that $\langle v,v\rangle>0$.

We first note that $v,w$ are linearly independent. Otherwise $w=kv$ for some $k\in \mathbb R$ and hence\[ \langle w,w\rangle = \langle kv,kv\rangle=k^2\langle v,v\rangle\geqslant 0.\]Then we consider a variable $t$ and the vector $tv+w$. We have\[\langle tv+w,tv+w\rangle=\langle  v,v\rangle t^2+2\langle v,w\rangle t+\langle w,w\rangle.\]The discriminant of this quadratic equation is\[\Delta=4(\langle v,w\rangle)^2-4\langle v,v\rangle\langle w,w\rangle>0\]since $\langle v,v\rangle >0$ and $\langle w,w\rangle < 0$. Therefore, we can find a real number $t_0$ such that $\langle t_0v+w,t_0v+w\rangle=0$. Then by Definiteness of the definition, we have  $t_0v+w=0$ which contradicts the fact that $v$ and $w$ are linearly independent.


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 $y_i=1$.


13. Solution: From the Law of Cosines, we have

$$ \begin{aligned} ||u – v||^2 &= ||u||^2 + ||v||^2 – 2 ||u||\:||v|| \cos \theta\\ ||u||^2 – 2 \langle u, v \rangle + ||v||^2 &= ||u||^2 + ||v||^2 – 2||u||\:||v|| \cos \theta\\ -2 \langle u, v \rangle &= -2 ||u||\:||v|| \cos \theta\\ \cos \theta &= \frac{\langle u, v \rangle}{||u||\:||v||}\\ \end{aligned} $$


14. Solution: The Cauchy-Schwarz Inequality ensures $\frac{\langle x, y \rangle}{||x||\:||y||}$ is always a number between $-1$ and $1$. Moreover it only equals $1$ or $-1$ if one of $x, y$ is a scalar multiple of the other, that is, when the angle is $0$ or $\pi$.


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.


23. Solution: We have $\langle (v_1, \dots, v_m), (v_1, \dots, v_m) \rangle = \langle v_1, v_1 \rangle + \dots + \langle v_m, v_m \rangle$. Positivity and definiteness follow directly from the fact that $\langle v_j, v_j \rangle \ge 0$ for each $j$. The remaining properties follow easily from the definitions.


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.


26. Solution:(a) We have

$$ \begin{aligned} \langle f(t), g(t) \rangle’ &= \lim_{h \to 0} \frac{\langle f(t+h), g(t+h) \rangle – \langle f(t), g(t) \rangle}{h}\\ &= \lim_{h \to 0} \frac{\langle f(t+h), g(t+h) \rangle – \langle f(t), g(t+h) \rangle + \langle f(t), g(t+h) \rangle – \langle f(t), g(t) \rangle}{h}\\ &= \lim_{h \to 0} \frac{\langle f(t+h) – f(t), g(t+h) \rangle + \langle f(t), g(t+h) – g(t) \rangle}{h}\\ &= \lim_{h \to 0} \frac{\langle f(t+h) – f(t), g(t+h) \rangle}{h} + \lim_{h \to 0} \frac{\langle f(t), g(t+h) – g(t) \rangle}{h}\\ &= \lim_{h \to 0} \langle \frac{f(t+h) – f(t)}{h}, g(t+h) \rangle + \lim_{h \to 0} \langle f(t), \frac{g(t+h) – g(t)}{h} \rangle\\ &= \langle f'(t), g(t) \rangle + \langle f(t), g'(t) \rangle\\ \end{aligned} $$

(b) We have

$$ \langle f(t), f(t) \rangle’ = \langle f'(t), f(t) \rangle + \langle f(t), f'(t) \rangle = 2\langle f'(t), f(t) \rangle. $$

However, $\langle f(t), f(t) \rangle’$ is $0$, because $\langle f(t), f(t) \rangle$ is constant (it equals $c^2$ for all $t$). Thus $\langle f'(t), f(t) \rangle = 0$.

(c) If $f(t)$ describes a trajectory on the surface of a sphere centered at the origin, then $f'(t)$ is perpendicular to $f(t)$.


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 \[\|w-u\|\le \|w-v\|\quad \text{for all } v\in C.\]


31. Solution: (From Marcel Ackermann)

Solution Manual Linear Algebra Done Right

6A31

The triangle can be extended to a parallelogram (rotation mirror at the intersection of d and c), see the picture above. Thus we can apply the parallelogram equality: $$2 (\Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2) = \Vert 2\vec{d} \Vert^2 + \Vert \vec{c} \Vert^2.$$ Therefore, we have $$\Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2 = \frac{1}{2} \Vert \vec{c} \Vert^2 + 2\Vert \vec{d} \Vert^2.$$

Linearity

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