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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 \[ \|u+av\|^2-\|u\|^2=|a|^2\|v\|^2+a\langle v,u\rangle +\bar a\langle u,v\rangle\ge 0. \]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 \[\|w-u\|\le \|w-v\|\quad \text{for all } v\in C.\]


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This Post Has 23 Comments

  1. It's very much a nit, but on the first line of the solution to #9, two 1's are missing after the inequality signs, i.e., the norm of u is less than or equal to 1 and the norm of v is less than or equal to 1.

  2. For #26, I only find it solvable for the Euclidean inner product. Does someone have an idea for other inner products?

  3. Does anyone have an idea of Q26(c)? It's just like an exercise of analysis, not algebra...

  4. Any one tried exercise 18?

    1. Use 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

    2. proof of the other side is easy

  5. For problem 17, we can simply let (x,y)=(-1,-1), then use positiveness of inner product

  6. Can you please solve 29b? I am lost. I would've solved it with orthonormal basis, but we haven't covered orthonormal basis yet.

    1. Nevermind. 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.

  7. does anyone have a proof for 6A30,I can't figure it out.

  8. For 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.

    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.

  9. Hi, 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. :)

  10. For 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?

    1. Please check the following $$a\langle v,u\rangle =\bar a\langle u,v\rangle=-\frac{|\langle v,u\rangle|^2}{\|v\|^2}.$$

      1. Yes, 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!

  11. I posted a solution to #3 at my blog.

    1. Great! This is the only answer that gets the right point. Thank you

  12. Solution 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: $2 (\Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2) = \Vert 2\vec{d} \Vert^2 + \Vert \vec{c} \Vert^2 \Leftrightarrow \Vert \vec{a} \Vert^2 + \Vert \vec{b} \Vert^2 = \frac{1}{2} \Vert \vec{c} \Vert^2 + \Vert \vec{d} \Vert^2$

  13. Does anybody have a proof for 6A5 for the infinite dimensional case? (eg. showing surjectivity)

  14. For problem 27, the given substitution does not work. Try a = w/2 - u/2 and b = w/2 - v/2 instead.

    1. Yes, you are correct. Thank you.

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