I'm studying to get into a Master's degree in Pure Mathematics in Brazil and I came across this question:
Let $V$ and $W$ vector spaces of arbitrary dimensions (finite or infinite). Show that exists $T : V \longrightarrow W$ injective $\Leftrightarrow $ dim V $\leq$ dim W.
This question has been asked here on the forum several times with the hypothesis that V and W are finite-dimensional vector spaces. I would like a proof that applies to both the finite case and the infinite case.
I managed to prove $\Rightarrow$. Could you give me tips on how to prove $\Leftarrow$?
My partial proof:
$\Rightarrow$ Let $T : V \longrightarrow W$ be a injective linear transformation. Suppose that
$$\text{dim} \ V \ > \ \text{dim} \ W,$$
and let $\alpha$ be a basis of $V$. How $T$ is injective, then $T(\alpha)$ is a linearly independent subset of $W$. Note that the subspace generated by $T(\alpha)$ is isomorphic to $V$ and is contained in $W$. This is absurdity because every subspace of W has the dimension less than equal to the dimension of W.
$\Leftarrow$ I wasn't able to.
Best Answer
There really is no difference between the finite and infinite cases. Choose bases $B, C$ of $V,W$ respectively. If $\dim(V)\leq\dim(W)$ then $|B|\leq |C|$, and so there is an injective map $f:B\to C\subseteq W$. Now just extend it to a linear map $V\to W$. It's easy to check that this extension is injective.
Conversely, if $T:V\to W$ is an injective linear map, and $B$ is a basis of $V$, then $T(B)$ is a linearly independent subset of $W$ of cardinality $|T(B)|=|B|$. As a linearly independent subset it can be extended to a basis of $W$. Hence, the cardinality of a basis of $W$ is at least $|B|=\dim(V)$.