# Prove that $L: V \rightarrow W$ is surjective if L is injective

Let $$L: V \rightarrow W$$ be an injective linear transformation.

Let $$dim V , dimW = n$$.

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...

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.
Take a basis $$v_1,\dots,v_n$$ of $$V$$, and show that $$L(v_1),\dots,L(v_n)$$ is a basis for $$W$$.