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I need to prove that for every Vector Space this is valid: $$ -v = (-1)*v $$

-v = inverse element of addition

-1 a real number

$*$ the multiplication by real number of the Vector Space

My teacher said that $-v$ is just a notation for the inverse element of addition. I'd like to prove that $-v = (-1)*v$.

So I came up with the following solution and I'd like to know if it's correct: \begin{align} v + (-1)*v = u \\ (1)*v + (-1)*v = u \\ (1-1)*v = u \\ 0*v = u \\ o = u\\ \end{align}

Since $v + -v = o$

$v + (-1)*v = o = v + -v$ adding -v to both sides v + (-1)*v -v = v + -v + -v o + (-1)*v = o + -v (-1)*v = -v

Did I commit any mistakes? Did I make any assumptions that may not be valid for EVERY Vector Space?

Edit: As S. Sheng said I have not proved that (0)*v = o. I'll try to prove that and come back later with a proof of that. I also haven't proved that (1)*v = v Oh my... I'm starting to think this is beyond my abilities..

leo
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Miranda
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  • Although it pertains to a somewhat different question, my answer at http://math.stackexchange.com/questions/944840/prove-that-if-u-w-v-w-then-u-v-without-additive-inverse/945052#945052 might be of help here. In particular, in any discussion of how to prove "obvious" properties of vectors, you really need to specify the axioms you're assuming. – Barry Cipra Sep 30 '14 at 16:32
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    The fact follows from the axioms. Since the axioms are common to all vector spaces, the identity holds for all vector spaces. – MathematicsStudent1122 Apr 08 '18 at 10:43
  • @Miranda Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 10 '18 at 12:47

7 Answers7

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One way to deduce it is $$ \vec0=0\vec u\\ \vec0=(1+(-1))\vec u\\ \vec0=1\vec u+(-1)\vec u\\ \vec 0+(-\vec u)=\vec u +(-1)\vec u+(-\vec u)\\ -\vec u=(\vec u +(-\vec u))+(-1)\vec u\\ -\vec u=(-1)\vec u $$

Arthur
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  • Your proof seems to be incomplete or, less likely, relying on an unconventional axioms of vector space. $\overrightarrow{0} = 0\overrightarrow{u}$ is not itself an axiom according to https://en.wikipedia.org/wiki/Vector_space. – Dmitrii Demenev May 06 '21 at 23:33
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The u is extraneous for your proof. But in any case the edited version is now correct. I will provide a rewrite so that others can use this thread for future reference.

$$v+(-1)v=(1)v+(-1)v=(1-1)v=(0)v=0$$

Adding $-v$ to both sides yields $(-1)v=-v$

Note that in our proof we assume $(0)v=0$. I am assuming that you have already proved that yourself or are allowed to assume it.

Enigma
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See the Axiom of the vector spaces

Axiom (4),(6) give the desired result. one more note:-In a group, the additive inverse is unique!

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$$(-1)\vec{u}+(\vec{u}) =(-1)\vec{u}+(1)(\vec{u})=$$

$$ (-1+1)(\vec{u})= 0(\vec{u})=0$$

That makes $$(-1)\vec{u}=-\vec{u}$$

Mohammad Riazi-Kermani
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In any vector space for any vector $\vec u$ exists an inverse element of addition $\vec v$ such that

$$\vec u+\vec v=\vec 0$$

that is precisely $\vec v=-\vec u=(-1)\vec u.$

user
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Let $v \in V$, where is $V$ is a vector space. $(-1)v = (-1)v + 0 = (-1)v + (v+(-v)) = (-1)v + v + (-v) = (-1+1)v + (-v) = -v$ You don't need to prove that $1v=v$ because it's an property of vector space. But you need to prove that $0v=0$, where is $0$ on the left side is scalar and vector is on the right side. $0v=(0+0)v = 0v+0v$ $0v+(-0v) = 0v+0v+(-0v)$ Thus we got that $0v = 0$. In both proofs i used only properties of vector space.

perepelart
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Instead of saying "Since $v +-v = 0$", you might want to say "Hence $v + (-1)v = 0$". And from this (depending on your definition, it follows immediately that $-v + (-1)v$.

Also, be sure to follow your teachers notation. He/she might want you to write $\vec{v}$ for vectors. That way we can distinguish between the number $0$ and the vector $\vec{0}$.

Also, if you insist on having the $u$ in your proof, then I would say in the beginning of the proof "Let $u = v + (-1)v$".

Thomas
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