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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?

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No, $R^4$ consists of points in 4 dimension while $R^2$ consists of points in 2 dimension. – NasuSama Feb 15 '14 at 3:30
The Rank-Nullity Theorem can help you prove this. Still, there are bijections between $R^2$ and $R^4$ , just not linear ones. – user99680 Feb 15 '14 at 3:31
To show that a function is onto, you need to show that for every $y \in R^4$, there exists $x \in R^2$ such that $T(x) = y$. As user99680 said, this theorem might work. – NasuSama Feb 15 '14 at 3:31
In what sense is $\mathbb R^4$ "greater than" $\mathbb R^2$? The two sets have the same cardinality... – Thomas Andrews Feb 15 '14 at 3:36
I think it is alright to say $ℝ^4$ is greater than $ℝ^2$, because $ℝ^2$ is a zero-set under any isometric embedding into $ℝ^4$. – canaaerus Feb 15 '14 at 9:56
up vote 2 down vote accepted

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 $1-1$.

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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.

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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.

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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.

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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.

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Did you mean "saying that $T$ is not surjective"? In any case, the last sentence is the key argument and there's no need to argue by contradiction. (I was not the downvote, but your second sentence is false as stated.) – Mark S. Feb 15 '14 at 3:43
@MarkS.: Thanks for telling me; what you wrote is what I had intended to write. I've fixed it. (Was the omission where the possibility of proof-by-contradiction came in? I wasn't trying to prove it that way.) – dmk Feb 15 '14 at 3:49
I think I just forgot what the original question was, so never mind that part. – Mark S. Feb 15 '14 at 3:52

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