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.
Daniel
13 Feb 2022Can 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
Quester
7 Oct 2022Isn't question 4 based on a wrong assumption? It claims that only one of the criteria for a vector space is not met under definition 1.19, but in addition to not having an additive identity, I would also claim that the empty set does not have an additive inverse.
Given that the additive inverse requires a set to contain two elements that add to 0, that would surely also be impossible with the empty set.
I assume that I'm wrong though, so would somebody please tell me why the empty set has an additive inverse given the definition under 1.19?
Sariel
11 Aug 2021Please explain Q6 more detailed
Jane
11 Aug 2021Could you explain Q6 more detailed？thanks
matizzat
30 Dec 2020In exercise 6 is it valid if I answer that (-infty) + (infty) is not defined and thus the conmutative property doesn't hold?
Linearity
5 May 2021No. Because if you multiply $\infty+(-\infty)=0$ by $-1$, you will get $(-\infty)+(\infty)=0$.
hfz223322
20 May 2021But the distributive properties need to be proved before you use it.
Linearity
24 May 2021Argue 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 2020For 5 we do not know multiplicative inverse just from 1.19.
Lost Ghost
3 Feb 2020Q4 is incorrect
Tianze
20 May 2020Are you saying that the title statement itself is wrong?
disqus_4MuuHb6gF1
22 Jun 2019The 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
Mohammad Rashidi
22 Jun 2019The proof is complete because if a=0, then it is already true..
Tianze
20 May 2020The proof of the two of you is equivalent, because there are only four cases to discuss
stan_comp_neuro
13 Feb 2019I 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
Mohammad Rashidi
22 Jun 2019It implies that if a!=0, then v=0. If a=0, then it is already true. The proof is complete.
Paweł Pod
9 Jul 2018An 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 2017Alternatively, we could just observe that $ t + \infty = \infty$ for every $ t \in \mathbf{R}$ automatically violates uniqueness of additive identity.
Luis
15 Oct 2020I thought the same
Linearity
5 May 2021This does not show $\infty$ is an additive identity.
Rong Ou
23 Apr 2017For 6 it also violates associativity:
(infty + infty) + (-infty) = infty + (-infty) = 0
infty + (infty + (-infty)) = infty + 0 = infty
MASHIAT MUTMAINNAH
18 Jul 2016Hi! 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!
Mohammad Rashidi
14 Jan 2017Since zero vector should be unique.
Abu-Yakitori
27 Feb 2017Why 0 + t = infinity + t?
spindash
28 Feb 2017From 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.