can the null space contain more elements that {0} if the map is a function can the null space have more elements than $\{0\}$ if T is a function but not necesarily linear?
For example, For the map 
$$T(a_1,a_2)\mapsto (\sin a_1,0)$$ would'nt it be the case that if $T$ were linear then only $T(0,0)$ maps to (0,0), so showing two distinct elements that map to (0,0) should prove non-linearlity?  I do have another way off proving that the map is'nt linear 
$$T(2\dfrac{\pi}{2},0)= (0,0)$$ but factoring out the $2$ maps $T$ to $(2,0)$.  So I'm just wondering if the first proof I wrote is sufficient?
Thanks in advance
 A: I think you are mixing up some concepts.
To clarify a bit: a linear mapping  $T: V \rightarrow W$ over a field $K$ satisfies the following conditions.  For all $u,v \in V$ and any scalar $\alpha \in K$ 


*

*$T(u+v) = T(u) + T(v)$

*$T(\alpha v) = \alpha T(v)$ 


Now, to your questions.

can the null space have more elements than {0} if T is a function
  but not necessarily linear?

Yes, there are many linear mappings that have more elements in the null space than just $\{ 0\}$. That's why the concepts of null spaces and ranges are so important in linear algebra.

would'nt it be the case that if T were linear then only $T(0,0)$maps to $(0,0)$, so showing two distinct elements that map to $(0,0)$ should prove non-linearlity?

No. If a linear mapping $T$ has more elements in the null space than just $\{ 0\}$ it just means that the mapping is not injective. It does not mean it's not linear.
A: If we stick with the commonly accepted meaning of common terms, then unfortunately your question doesn't make much sense.
First of all, if $T$ is a function, even between vector spaces, that is not linear, then we can't speak of its "null space". You're actually talking about the preimage of zero: for $T:V\to W$, you seem to be asking about $T^{-1}(0_W)=\{v\in V:T(v)=0_W\}$. This is guaranteed to be a vector subspace of $V$ only if $T$ is linear, in which case it is known as the null space or the kernel of $T$. But if $T$ is not linear, then it's usually NOT a subspace, just a subset, and so we wouldn't call it "the null space".
Moreover, for general functions the preimage of zero may contain not only more than just $\{0\}$, but also less than $\{0\}$, or something completely different at all. For example, consider the following nonlinear functions $T:\mathbb{R}\to\mathbb{R}$ between one-dimensional real vector spaces:


*

*$T(x)=x^2+1$, where $T^{-1}(0)=\varnothing$ is "less" than $\{0\}$;

*$T(x)=x^2-5x+6$, where $T^{-1}(0)=\{2,3\}$ is "more" (?) than just $\{0\}$, except for it doesn't even contain zero.

