Null space helping to know whether a linear transformation is one one or not A linear transformation is one one iff its null space contains only the zero vector.
I have understood the proof of it,but I wasn't able to get the flavour of the result,I mean how is null space helping us to determine one one?I am not able to see the connection.
Any Intuition behind the result?
 A: Perhaps this is a helpful perspective.
Suppose that $A:V \to W$ is a transformation between vector spaces. The statement that $A$ is one-to-one is equivalent to saying that for each $w \in W$, the "linear system"
$$
A(v) = w
$$
has at most one solution for each $w \in W$. On the other hand, the null space of $A$ (which I denote $\ker A$) is important to describing the set of solutions to such an equation. If $A(v) = w$ has a solution $v = v_p$ (i.e. $v_p$ is such that $A(v_p) = w$), then the solution set can be written as
$$
S = \{v_p + v_n : v_n \in \ker(A)\}.
$$
With that, we can see what's happening. $A$ is one-to-one if and only if $S$ contains at most one element for each $w$, which only occurs if $\ker(A)$ contains exactly one element.
A: The intuition is this:

The preimage of each point in the image is a copy the nullspace.

If the nullspace is zero-dimensional, then each point in the image has exactly one preimage. If the nullspace is not zero-dimensional, then each point in the image has a whole (affine) subspace of preimages.

Formalizing this intuition: Let $T:V\to W$ be a linear transformation with nullspace $\operatorname{Null}(T)$. Let $w\in W$ be in the image of $T$. Then the preimage of $w$ is
$$ T^{-1}(w) = \{ v\in V\ |\ Tv = w\}$$
Any two points $v_1,v_2\in T^{-1}(w)$ must have $Tv_1 - Tv_2 = 0$, ie, $v_1 - v_2\in \operatorname{Null}(T)$. Conversely, if $x\in \operatorname{Null}(T)$ and $v\in T^{-1}(w)$, then $v+x \in T^{-1}(w)$ because $T(v+x) = Tv + Tx = Tv = w$.
So, given any one $v_0\in T^{-1}(w)$, we can find the entire preimage:
$$ T^{-1}(w) = v_0 + \operatorname{Null}(T) $$
