Use properties of definition of a vector space to prove that $0v = \mathbf{0}$ for any $v$

Question

My proof is:

There exists $-v$ such that $\mathbf{0} = -v + v = \mathbf{0}$; Then $\mathbf{0} = -v + v = [(-1)+1] v = 0 v$.

Is this proof correct?

Answer 1

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.

Answer 2

( 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