Let $L: V \rightarrow W$ be an injective linear transformation.
Let $\dim V , \dim W = n$.
Show that $L$ is surjective.
My thoughts:
If $L(V)$ is the image of $V$, we can show that that $L: V \rightarrow L(V)$ is a bijection and thus that $\dim L(V) = n$. And then from there maybe you can show that $L(V)$ has to be equal to $W$ (surjective) maybe with a proof by contradiction but I can't quite make it work…
Best Answer
You have the right idea. Any $n$-dimensional subspace of an $n$-dimensional vector space must be the whole space. Why? Well, assume you have $w \in W$ that is not in $L(V)$. Then you can add $w$ to a basis of $L(v)$ to form a new subset of $W$. However, this subset now has $n+1$ linearly independent vectors, a contradiction.