# Use properties of definition of a vector space to prove that $0v = 0$ for any $v$

My proof is:

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

• Yes, your proof is correct. Oct 26, 2016 at 22:55

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.

• Thx for another proof. Oct 26, 2016 at 22:58
• 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. Oct 26, 2016 at 23:00
• Could you elaborate a little more on those situations? Oct 27, 2016 at 1:40

( 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