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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.


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This Post Has 25 Comments

  1. 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

    1. Isn'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?

  2. Please explain Q6 more detailed

  3. Could you explain Q6 more detailed?thanks

  4. In exercise 6 is it valid if I answer that (-infty) + (infty) is not defined and thus the conmutative property doesn't hold?

    1. No. Because if you multiply $\infty+(-\infty)=0$ by $-1$, you will get $(-\infty)+(\infty)=0$.

      1. But the distributive properties need to be proved before you use it.

        1. Argue it by contradiction. Namely, first assume it is a vector space and then show something is impossible. So you can use distribution law.

  5. For 5 we do not know multiplicative inverse just from 1.19.

  6. Q4 is incorrect

    1. Are you saying that the title statement itself is wrong?

  7. 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. The proof is complete because if a=0, then it is already true..

    2. The proof of the two of you is equivalent, because there are only four cases to discuss

  8. 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. It implies that if a!=0, then v=0. If a=0, then it is already true. The proof is complete.

  9. 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.

  10. Alternatively, we could just observe that $ t + \infty = \infty$ for every $ t \in \mathbf{R}$ automatically violates uniqueness of additive identity.

    1. I thought the same

    2. This does not show $\infty$ is an additive identity.

  11. For 6 it also violates associativity:

    (infty + infty) + (-infty) = infty + (-infty) = 0
    infty + (infty + (-infty)) = infty + 0 = infty

  12. 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. Since zero vector should be unique.

      1. Why 0 + t = infinity + t?

        1. 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.

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