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 \[ \|u+av\|^2-\|u\|^2=|a|^2\|v\|^2+a\langle v,u\rangle +\bar a\langle u,v\rangle\ge 0. \]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 \[\|w-u\|\le \|w-v\|\quad \text{for all } v\in C.\]
Evan
16 Aug 2021For #26, I only find it solvable for the Euclidean inner product. Does someone have an idea for other inner products?
Evan
14 Aug 2021Does anyone have an idea of Q26(c)? It's just like an exercise of analysis, not algebra...
Xinyu
9 Dec 2020Any one tried exercise 18?
Allen
7 Feb 2021I found the question answered by Lukas Geyer at
https://math.stackexchange.com/questions/1331862/prove-that-there-is-an-inner-product-on-mathbbr2-given-that-the-associate.
Evan
14 Aug 2021Use u=(1,0) and v=(0,1) while testing this inner product with Parallelogram Equality (6.22). You will find p=2 is the only solution to the equation
Evan
14 Aug 2021proof of the other side is easy
Xinyu
5 Dec 2020For problem 17, we can simply let (x,y)=(-1,-1), then use positiveness of inner product
Phi
24 Jun 2020Can you please solve 29b? I am lost. I would've solved it with orthonormal basis, but we haven't covered orthonormal basis yet.
Phi
25 Jun 2020Nevermind. I asked Axler what the intended solution is. He replied that the intended solution uses the existence of an orthonormal basis, and that he included the problem 29b in section A by mistake - it should be moved to section B.
katharine
2 Jun 2020does anyone have a proof for 6A30,I can't figure it out.
Charlie Fan
1 Jun 2020For Q3, it basically requires that you prove the following statement:
if is greater than 0 for some v, then this inequality holds for all non-zero v's.
Proof:
Now suppose > 0 and < 0, then we construct a quadratic function of t:
f(t) = , where t is a real number
It's easy to show that f(0) > 0 and f(t) < 0 when t is large, then by the continuity of quadratic function there exists a t0 such that
= 0
By part (b) of the definition of inner product this is equivalent to
u + t0*v = 0
therefore u = -t0*v
Then you can use the homogeneity property to show that > 0, a contradiction.
hteica
6 May 2020Hi, thanks for the great solutions. I think there are some errors in the solution n.6, since = ||v||^2, I think you missed the ^2 there, it should've been |a|^2 * ||v||^2. I would suggest to work with v/||v|| =: v', so ||v'|| = 1 and things will be less messy that way. :)
Zheng Chen
25 Jan 2020For Q6, I cannot understand the part "plug in a =.......and we will obtain.......". How to calculate it? Do we assume a is real? How do we deal with the 'a conjugate' then?
Linearity
25 Jan 2020Please check the following $$a\langle v,u\rangle =\bar a\langle u,v\rangle=-\frac{|\langle v,u\rangle|^2}{\|v\|^2}.$$
Zheng Chen
26 Jan 2020Yes, I think I got it, but I think an easier way to understand is to first use the orthogonal decomposition on u. Write u = cv + w so we have = 0, then we substitute all u with cv + w and expand it.
Again, thank you very much!
Mathily
18 Aug 2017I posted a solution to #3 at my blog.
Jiawei Wu
4 Oct 2020Great! This is the only answer that gets the right point. Thank you
Marcel Ackermann
19 Jun 2017Solution to 6A31:
The triangle can be extended to a parallelogram (rotation mirror at the intersection of d and c). Thus we can apply the parallelogram equality: $2 (\Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2) = \Vert 2\vec{d} \Vert^2 + \Vert \vec{c} \Vert^2 \Leftrightarrow \Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2 = \frac{1}{2} \Vert \vec{c} \Vert^2 + \Vert \vec{d} \Vert^2$
Marcel Ackermann
23 Apr 2017Does anybody have a proof for 6A5 for the infinite dimensional case? (eg. showing surjectivity)
Brian Lubeck
9 Feb 2019Per the errata: Page 175, Exercise 5: Assume that V is finite-dimensional. http://linear.axler.net/LADRErrataThird.html
Dalton Burke
31 Mar 2017For problem 27, the given substitution does not work. Try a = w/2 - u/2 and b = w/2 - v/2 instead.
Mohammad Rashidi
27 Jul 2017Yes, you are correct. Thank you.