suppose$x\in V$ , $V$ is a vector space. if $3x=0$, then $x=0$ suppose$x\in V$ , $V$ is a vector space. if $3x=0$, then $x=0$ It seems very trivial for me but i am not so sure how it works in vector space
 A: You have a vector space $V$ that is defined over a field; for simplicity, assume your field is the real numbers. Thus you can do scalar multiplication: you can multiply any vector by any real number. 
Every element of a field (except 0) has an inverse. Thus if $a \neq 0$ and $x \in V$ then if we have $ax=0$ we can multiply both sides by the inverse of $a$, $a^{-1}$, to get $x=0$. I.e., $a^{-1}ax = a^{-1}0 \implies 1x = 0 \implies x=0$.
In your particular case we have $$3x=0 \implies\frac{1}{3}3x = \frac{1}{3}0 \implies x=0.$$
EDIT: the fact that any scalar times the zero vector gives the zero vector again is a consequence of the vector space axioms: if $a$ is a scalar and $u$ and $v$ are vectors then we know that $a(u+v) = au + av$. If we set $v$ to be the zero vector we get
$$
  au =a(u) = a(u+0) = au+a0 \implies au = au+a0 \implies a0 = 0
$$
A: If you are writting $3x = 0$ that means that $0 \in V$.
With that given you can always multiply both sides by the same scalar, in this case $1/3$. The 0 vector will remain the same and $x = 0$. So $x$ is the null vector of the space.
So your reasoning is good and trivial using the properties of vector spaces.
A: That one is an special case of the following:
If $V$ is a real vector space, $x,y\in V$ and $t\ne 0$, then $tx=ty \Rightarrow x=y$.
