How to intuitively explain that if kernel(L) equal to the set containing the zero vector, then L is one-to-one? Whenever I try to learn about a relationship, I try to reason intuitively why a theorem or lemma should make sense. I know often times this is increasingly difficult to achieve. However, I have the following:
Let $ L: \mathbb{V} \to \mathbb{W} $ be a linear mapping. $L$ is one-to-one if and only if Ker($L$) = $\{\vec{0}\}$
I can intuitively explain the forward direction, that if $L$ is one-to-one, then only one value can possibly map to $\vec{0}$ and that must be $\vec0$. 
When I consider the other direction, it is hard for me to intuitively explain that the property of a linear mapping being one-to-one follows naturally from the fact that its kernel contains only $\vec0$. I know that the algebraic proof of this is trivial, but I feel like without a more intuitive understanding, I am misunderstanding something about linear mappings and hence not fully appreciating the importance of this relationship.
Thanks in advance.
EDIT:
The question is, can I explain the logic behind this principle without implicitly referring to the proof?

We know that the kernel is a subvectorspace. Hence if $x_1, x_2$ are (two different elements, nonzero) in the kernel, we have that $x_1 - x_2$ is in the kernel, so we find that
$$0 = L(x_1 - x_2) = L(x_1) - L(x_2)$$
by linearity. This would mean that $L$ is not one-to-one, since we have two different points with the same image.
I think this is just such a basic concept, that any 'non algebraic argument' can be translated to an algebraic argument en vice versa. I could translate my algebraic explanation into words as follows: If we suppose that the map is not injective, then there are different elements having the same image, hence by linearity their difference is mapped onto zero. Therefore the kernel is not trivial. But this is essentially what I wrote above.
