Existence of a Surjective Linear Map $\mathbb{R}^3$ to $\mathbb{R}^4$? Is it possible for there to be a surjective linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^4$?
I feel like this isn't possible because $\mathbb{R}^3$ has fewer vectors than $\mathbb{R}^4$. If so, linear transformations between the two spaces should be injective. But if that's the case, for all linear transformations $\mathbb{R}^3\to\mathbb{R}^4$, the kernel should only ever contain the zero vector. But this is not the case. Why?
 A: There are functions from $\Bbb R^3$ to $\Bbb R^4$ which are surjective, but none of them preserve the vector space structures. By the rank-nullity formula: $$3 = \dim \ker T + \dim {\rm Im } \ T,$$ so we can't have $\dim {\rm Im } \ T = 4 $.
A: No. In $\mathbb R^4$ there are linearly independent sets of $4$ vectors. This is a problem, since if $f$ is a linear map and $S$ is a set of vectors, then we can prove:

If $f[S]$ is linearly independent, then $S$ is.

which is easy to do by looking at the contrapositive that if $S$ is dependent than so is $f[S]$, which is obvious since if $\alpha_1 s_1 +\alpha_2 s_2 +\ldots+ \alpha_n s_n=0$, then so $f(\alpha_1 s_1 +\alpha_2 s_2 +\ldots +\alpha_n s_n)=0$, but we can expand the left hand side to
$$\alpha_1f(s_1)+\alpha_2f(s_2)+\ldots+\alpha_n f(s_n)=0$$
implying a dependency in $f[S]$.
If $f:\mathbb R^3 \rightarrow \mathbb R^4$ were surjective, then we could take some set $V$ of $4$ linearly independent vectors in $\mathbb R^4$ and find some set of four vectors $S$ in $\mathbb R^3$ such that $f[S]=V$. However, this is impossible, since it would mean that the vectors of $S$ are linearly independent, but this can't happen in $\mathbb R^3$, which has dimension $3$.
A: No, the dimension of the range of such a linear transformation is at most $3$, by the rank-nullity theorem.
A: A linear function whose domain is $\mathbb R^3$ is determined by its values at a basis of $\mathbb R^3$, which contains just three vectors.  The image of a linear map from $\mathbb R^3$ to $\mathbb R^4$ is the span of a set of three vectors in $\mathbb R^4$, and the span of only three vectors is less than all of $\mathbb R^4$.
However, that is no reason why the map should be injective.  The three vectors in $\mathbb R^4$ that are the images of the aforementioned basis vectors in $\mathbb R^3$ need not be linearly independent.  If they are linearly independent, then the map is injective; otherwise it is not.
A: They have the same cardinality but I don't believe cardinality implies existence of surjective linear transformation.
