Chapter 1 Exercise B

1. Solution: By definition, we have\[(-v)+(-(-v))=0\quad\text{and}\quad v+(-v)=0.\]This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.

2. Solution: If $a=0$, we are done.

If $a\ne 0$, then $a$ has inverse $a^{-1}$ such that $a^{-1}a=1$. Hence \[v=1\cdot v=(a^{-1}a)v=a^{-1}(av)=a^{-1}\cdot 0=0.\]Here we use associativity in 1.19 and and 1.30.

3. Solution: Let $x=\dfrac{1}{3}(w-v)$, then \[v+3x=v+3\cdot \dfrac{1}{3}(w-v)=v+(w-v)=w.\]This shows existence. Now we show uniqueness. Suppose, we have another vector $x’$ such that $v+3x’=w$. Then $v+3x’=w$ implies $3x’=w-v$. Similarly, $3x=w-v$. Hence \[3(x-x’)=3x-3x’=(w-v)-(w-v)=0.\]By Problem 2, we must have $x-x’=0$, that is $x=x'$. This shows uniqueness.

4. Solution: Additive identity: there exists an element $0\in V$ such that $v+0=v$ for all $v\in V$ ; This means $V$ cannot be empty.

5. Solution: If we assume the additive inverse condition, we already showed $0v=0$ for all $v\in V$ in 1.29. Now we assume $0v=0$ for all $v\in V$ and then show additive inverse condition. Since we have $0v=0$ for all $v\in V$, we have \[ v+((-1)v)=1v+((-1)v)=(1+(-1))v=0v=0, \]this means the existence of additive inverse, i.e. the additive inverse condition.

6. Solution: This is not a vector space over $\mathbb R$. Consider the distributive properties in 1.19. If this is a vector space over $\mathbb R$, we will have \[\infty=(2+(-1))\infty=2\infty+(-1)\infty=\infty+(-\infty)=0.\]Hence for any $t\in\mathbb R$, one has\[t=0+t=\infty+t=\infty=0.\]We get a contradiction since zero vector is unique.

This Post Has 24 Comments

Daniel
13 Feb 2022 Reply
Can also use associativity of scalar multiplication twice, in addition to the multiplicative identity.
-(-v) = (-1) * (-v) = (-1) * ( (-1)*v ) = ( (-1)*(-1) ) * v = (-1)^2 * v = 1 * v = v
Sariel
11 Aug 2021 Reply
Please explain Q6 more detailed
Jane
11 Aug 2021 Reply
Could you explain Q6 more detailed？thanks
matizzat
30 Dec 2020 Reply
In exercise 6 is it valid if I answer that (-infty) + (infty) is not defined and thus the conmutative property doesn't hold?
1. Linearity
  5 May 2021 Reply
  No. Because if you multiply $\infty+(-\infty)=0$ by $-1$, you will get $(-\infty)+(\infty)=0$.
  1. hfz223322
    20 May 2021 Reply
    But the distributive properties need to be proved before you use it.
    1. Linearity
      24 May 2021 Reply
      Argue it by contradiction. Namely, first assume it is a vector space and then show something is impossible. So you can use distribution law.
Listening
15 Oct 2020 Reply
For 5 we do not know multiplicative inverse just from 1.19.
Lost Ghost
3 Feb 2020 Reply
Q4 is incorrect
1. Tianze
  20 May 2020 Reply
  Are you saying that the title statement itself is wrong?
disqus_4MuuHb6gF1
22 Jun 2019 Reply
The proof of 2 is not complete. See alternative proof below:
Suppose av = 0, with a != 0 and v != 0
Since a != 0, there exists (1/a) in F such that (1/a)a = 1.
So we have:
av = 0
(1/a)av = (1/a)0
((1/a)(a))v = 0 ( by multiplicative associativity and 1.30 respectively)
(1)v = 0
v = 0, contradiction
Hence we conclude that a = 0 or v = 0
1. Mohammad Rashidi
  22 Jun 2019 Reply
  The proof is complete because if a=0, then it is already true..
2. Tianze
  20 May 2020 Reply
  The proof of the two of you is equivalent, because there are only four cases to discuss
stan_comp_neuro
13 Feb 2019 Reply
I think here gives a better answer for question 2, here you did not address why the a!=0 and V != 0 does not exist stackexchange/2595465
1. Mohammad Rashidi
  22 Jun 2019 Reply
  It implies that if a!=0, then v=0. If a=0, then it is already true. The proof is complete.
Paweł Pod
9 Jul 2018 Reply
An another argument:
8 = 8 + 0 = 8 + (8 + (-8)) = (8 + 8) + (-8) = 8 + (-8) = 8 + ((-8) + (-8)) = (8 + (-8)) + (-8) = 0 + (-8) = (-8), this is the contradiction with the assumption that 8 and (-8) denote distinct objects.
Maxis Jaisi
25 Apr 2017 Reply
Alternatively, we could just observe that $ t + \infty = \infty$ for every $ t \in \mathbf{R}$ automatically violates uniqueness of additive identity.
1. Luis
  15 Oct 2020 Reply
  I thought the same
2. Linearity
  5 May 2021 Reply
  This does not show $\infty$ is an additive identity.
Rong Ou
23 Apr 2017 Reply
For 6 it also violates associativity:
(infty + infty) + (-infty) = infty + (-infty) = 0
infty + (infty + (-infty)) = infty + 0 = infty
MASHIAT MUTMAINNAH
18 Jul 2016 Reply
Hi! Thank you so much for all the solutions, I really appreciate your hard work. Can someone please explain to me the contradiction in the solution Ex 1B question 6?
I just can't seem to understand it. Thank you!
1. Mohammad Rashidi
  14 Jan 2017 Reply
  Since zero vector should be unique.
  1. Abu-Yakitori
    27 Feb 2017 Reply
    Why 0 + t = infinity + t?
    1. spindash
      28 Feb 2017 Reply
      From the first equation we obtain ∞ = 0 (only using distributive properties).
      Whereas in the second equation we get t = 0
      That is the contradiction, 0 is not unique.
      0+t = ∞+t since by the result of the first equation ∞ = 0. So if 0=∞, 0+t=∞+t.