Theorem

${latex.inline0v = 0\quad \forall v \in V} Note that on the left side of the equation we have the scalar 0 and on the right side of the equation we have the vector 0.

Proof

For \({latex.inline[v \in V](v \in V)} we have \){latex.inline0v = (0 + 0)v = 0v + 0v}. We can add the inverse of 0v to both sides and get 0(vector) = 0v.

Note that the result here involves the additive identity of V(0 vector) and scalar multiplication. The only way we can sort of link those two together in V is the use the distributive property. That’s why we brought that in here.