If $g:V \rightarrow V$ is an injective linear transformation. Prove if $V$ is finite dimensional then $g$ is surjective.

I am asked to prove this without the rank nullity theorem

My Attempt at a Proof

For the $\implies$;
If $g:V \rightarrow V$ is injective then the dimension of the kernel is 0, and so as ($im$) and ($ker$) are both subspaces of $V$, $dim(im)=dim(v)$ which implies $g$ is surjective.

For the $\impliedby$ ;
If $g:V \rightarrow V$ is surjective, then $dim(v)=dim(im)$ and $dim(ker)$ must be 0, so $g$ is injective.

It seems to me that i have indirectly used the rank nullity theorem, would there be a more suitable proof that doesn't incorporate it?

Solution 1:

Let $v_1,\ldots,v_n$ be a basis for $V$ (so $V$ has dimension $n$), then since $g$ is injective, $g(v_1),\ldots,g(v_n)$ is linearly independent. However, it's a linearly independent set of $n$ vectors, and $V$ has dimension $n$, so it's a basis, hence the image of $g$ contains the span of $g(v_1),\ldots,g(v_n)$, which being a basis, is all of $V$.