Let two linear maps $T$ and $S$, both go from $V$ to $W$, are linear maps, if $\text{null } T = \text{null } S$ , then $T=S$. Let $T:V \rightarrow W$ and $S:V \rightarrow W$ be two linear maps. If $\text{null } T = \text{null } S$, then $T = S$.
Counter-Example: Let $a,b \in V$, consider $T:$ defined as $T(a)=0$ and $T(b)=1$ and $S:$ defined as $S(a)=0$ and $S(b)=2$, where $0,1,2 \in W$. 
Since $\text{null } T = \text{null } S$, but $T\neq S$.
Is this example sufficient to disprove the problem? 
 A: No, your example is not good. You didn't define any linear transformations. You just mentioned what will be the images of some two vectors in an unknown space. How do you know $T$ and $S$ don't have some other elements in the kernel? Maybe their null spaces are very different if you look at their general definitions. 
Anyway, here is a trivial conterexample. Define $T,S:\mathbb{R^n}\to\mathbb{R^n}$ by $T(x)=x$ and $S(x)=-x$. 
A: Your idea is (almost) correct, however your argument is wrong.
Let me write your argument as an idea first and then turn it into a real proof.
Idea: If we manage to find a vector space $V$, two vectors $a,b\in V$ and maps $T:V\rightarrow\mathbb{R}$ and $S:V\rightarrow \mathbb{R}$ such that $T$ sends only $\text{span}(a)$ to zero, $T(b)=1$ and $S$ sends only $\text{span}(a)$ to zero and $S(b)=2$. (Note that if $T(a)=0$ then $T(\lambda a)=0$, hence $T(a)=0$ is equivalent to sending $\text{span}(a)$ to zero). Then we find a counter-example.
Now we turn this idea into a formal proof: we use the fact that a linear map is completely determined by the values it assigns to a basis. We choose $V$ to be our favorite $2$-dimensional vector space, say $V=\mathbb{R}^2$ and we take $a,b$ to be a basis, say $a=(1,0),b=(0,1)$.
Then we find the (unique) linear map $T$ which sends $a$ to zero (and so the span of $a$ as well) and $b$ to $1$. This map is given by $T(x,y) = y$. [exercise: explain to yourself why there are no more linear maps with that property].
Similarly we define $S(x,y)=2y$ so $S$ sends $a$ to zero and $b$ to $2$.
Then $\text{null}(T) =\text{null}(S)= \text{span}(a) = \{(x,0):x\in\mathbb{R}\}$ however $T\not = S$ because $T(b)=1\not = 2=S(b)$.
A: As others have noted, the hypothesis can't possibly be correct.  If $T$ is a linear map and $\alpha$ is any non-zero scalar other than $1$, then $\alpha T$ is also a linear map and $\operatorname{ker} T = \operatorname{ker} \alpha T$, but $T \neq \alpha T$.
