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My proof is:

There exists $-v$ such that $-v + v = 0$; Then $-v + v = [(-1)+1] v = 0 v$. Is it right?

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    $\begingroup$ Yes, your proof is correct. $\endgroup$ Oct 26, 2016 at 22:55

2 Answers 2

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Here's another proof:

Observe \begin{align} 0\cdot v = (0+0)\cdot v = 0\cdot v + 0\cdot v. \end{align} Subtracting from both sides yields \begin{align} \mathbf{0} = 0\cdot v - 0\cdot v = 0\cdot v+0\cdot v - 0\cdot v = 0\cdot v + \mathbf{0} = 0\cdot v. \end{align}

Note: I have used $\mathbf{0}$ to denote the zero element in the vector space and $0$ to denote the zero element in the scalar field.

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  • $\begingroup$ Thx for another proof. $\endgroup$
    – Eric Chen
    Oct 26, 2016 at 22:58
  • $\begingroup$ It should be noted that sometimes you might not have additive inverse for a given algebraic structure, i.e. not every $v$ has a $-v$ . But you still have a zero element, which then the above proof will work. $\endgroup$ Oct 26, 2016 at 23:00
  • $\begingroup$ Could you elaborate a little more on those situations? $\endgroup$
    – Eric Chen
    Oct 27, 2016 at 1:40
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( Here ⋅ denotes multiplication , + denotes addition operation. ) We can proof it by a simple equation.. $(1+0)\cdot v= 1\cdot v + 0\cdot v $ and $1\cdot v = v $

by definition in the vector space. $(1+0)\cdot v=1\cdot v=v $ then $v=v+0\cdot v$ Again in the vector space definition $0(\text{vector})+ v = v$ Then $0\cdot v $ must be zero vector

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