Implication when the product of scalar and a vector is the zero vector? For any scalar $s$ and vector $v$ ,it holds that $sv=0$ if and only if $s=0 \  or \ v=\vec0$.
A Textbook of linear Algebra  gives the following proof (the reverse direction is easy to prove):
Suppose that $sv=\vec0 $ and  $s \neq 0 $ then $$\ v=(s^{-1}s)\ v=s^{-1}(sv)=s^{-1}\vec 0=\vec 0 $$
I'm wondering if there is another approach that can be used to prove the forward direction.Any ideas?
(for example, by using the facts that  $sv=\vec0 $ and  $v \neq 0 $ can we prove that $s=0$ or can we  show the contrapositive.....  ..) ?
Is there another proof for the forward direction or no?

Holding for vector spaces with an inner product (inner product spaces):
Suppose $\vec{v}\neq \vec{0}$ and $s\vec{v}=\vec{0}$. Then
$$
0 = \frac{0}{\lVert \vec{v}\rVert^2} = \frac{\langle \vec{0}, \vec{v}\rangle}{\lVert \vec{v}\rVert^2} = \frac{\langle s\vec{v}, \vec{v} \rangle}{\lVert \vec{v}\rVert^2} = s\frac{\langle \vec{v}, \vec{v} \rangle}{\lVert \vec{v}\rVert^2} = s
$$
where we used the fact that $\lVert \vec{v}\rVert^2\neq 0$ (by assumption).
