Showing something is not onto? Quick question..:
If I have a linear transformation  $T:\mathbb R^2 \to \mathbb R^4$, can I say that since $\mathbb R^4$ is greater than $\mathbb R^2$, it can never be onto - since the elements in $\mathbb R^2$ will never be able to map to all the elements in $\mathbb R^4$?
Does this make sense?
 A: No, that reasoning alone is faulty. Surprisingly, perhaps, there are definitely bijections between $\mathbb{R}^m$ and $\mathbb{R}^n$ for any positive $m,n$. But you haven't used the fact that this is supposed to be a linear map.
A: Unfortunately, no. There are functions from $\mathbb R^2$ to $\mathbb R^4$ that are onto. For inspiration, see the wikipedia article for space-filling curves. 
You'll need to use a theorem from your notes/book/lecture about linear transformations to explain why that craziness can't happen with a linear transformation.
A: The rank-nullity theorem shows that a linear map $L :\mathbb R^k \rightarrow \mathbb R^n$  cannot be onto unless $k \geq n$. From Rank-Nullity, we have $$Dim(\mathbb R^2)=2 =DimKer(L)+Dim(Im(L))$$, so that the best you can get is$ Dim(Im(L))=2 < Dim(\mathbb R^4)=4 $, when $Dim(Ker(L)=0$ , i.e., when $L$ is $1-1$.
A: Funny, we did this in class today. Saying that $T$ is not surjective is equivalent to saying that $\text{range}(T) \subset \mathbb{R}^4$  that is, it's a proper subset. This in turn is equivalent to saying $\dim(\text{range}(T)) < \dim (\mathbb{R}^4)$. Using the fact that $\dim(\mathbb{R}^2) = \dim(\text{null}(T)) + \dim(\text{range}(T))$, we can get an upper bound for the dimension of the range.
A: The issue is largely about what you mean by "greater than", when you say $\mathbb{R}^4$ is "greater than" $\mathbb{R}^2$. Their "size" as sets, or cardinality, is actually equal, so there's no problem with the existence of a bijective function from $\mathbb{R}^2$ to $\mathbb{R}^4$.
Once we suppose that the function is linear, however, there is an issue, and the rank-nullity theorem is exactly what you need, as others have noted. What makes it work for you is that $\mathbb{R}^4$ is "greater than" $\mathbb{R}^2$ in terms of dimension as a vector space.
