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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.

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up vote 4 down vote accepted

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.

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so from $(ii)$ and $(iii)$ I know $dimV/Kerf=dim$ $Imf=dimV-dimKerf$. Thus $dimV-dimKerf=dim$ $Imf$ $\Longleftrightarrow dimV=dimKerf+dim$ $Imf$. $f$ is injective => $dimKerf=0$ => $dimV=dim$ $Imf$ => f surjective. | $f$ is surjective => $dim$ $Imf=dimV$ => $dimKerf=0$ => $f$ injective? – Tilman Dec 8 '13 at 1:54
Your proof is correct for the finite dimensional case (and also the last two sentences are correct), the problem in the infinite dimensional case is that it makes no sense to write $\mathrm{dim} \ V-\mathrm{dim} \ \mathrm{Ker} \ f$ as both could be infinite! Show that if $(u_i)$ is a basis of the image, $(w_j)$ a basis of the kernel, and $v_i$ with $f(v_i)=u_i$ then $(v_i,w_j)$ is a basis for $V$. – Alexander Grothendieck Dec 8 '13 at 11:50

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