Maybe I am not good at looking for the right questions but I haven't seen this task anywhere so I hope it is no duplicate.
I have to prove the following statement:
Let $V$ be a finite dimensional vector space and $f:V \rightarrow V$ a linear map. Show that $f$ is injective $\Longleftrightarrow$ $f$ is surjective.
The following is already known:
$(i)$ $\ker f$ and $\def\Im{\operatorname{Im}}\Im f$ are linear subspaces
$(ii)$ There exists an isomorphism $V/\ker f\rightarrow \Im f$
$(iii)$ If $U\subset V$ is a linear subspace then $\dim V/U=\dim V-\dim U$
$(iv)$ $f$ is injective $\Longleftrightarrow$ $\ker f=\{0\}$
So my approach is the following:
$"\Longrightarrow "$
Let $f:V\rightarrow V$ be injective. Then due to $(iv)$ $\dim\ker f=0$. Hence due to $(ii)$ and $(iii)$ $\dim V/\ker f=\dim V=\dim\Im f$. Since $\dim V=\dim\Im f$ it implies that $f$ is surjective.
$"\Longleftarrow "$
Let $f$ be surjective. Then $\dim\Im f=\dim V$. Additionally, because of $(ii)$ it is $\dim\Im f=\dim V/\ker f$. Thus $\dim V=\dim V/\ker f=\dim V-\dim\ker f$. This means that $\dim\ker f=0$ and hence $\ker f=\{0\}$ and thus $f:V\rightarrow V$ is injective.
For me it all makes sense but maybe I missed something.
Thank you in advance.
Best Answer
The proof is correct; personally I like to view it as a corollary of the rank theorem: $\mathrm{dim} \ V=\mathrm{dim \ Im} \ f+\mathrm{dim \ Ker} \ f$ (try to prove it, it is a generalization of your (iii); for the finite dimensional case it is immediate from (ii) and (iii)), which in fact holds for vector spaces of arbitrary dimension.