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

## 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.

## 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.