# Linear map $f:V\rightarrow V$ injective $\Longleftrightarrow$ surjective

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.

$$(i)$$ $$\ker f$$ 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.

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.
• 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? – user114193 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
• The poster correctly limited the statement of the theorem to the finite-dimensional case. It’s not true in the infinite dimensional case. Let $V$ be the space of sequences of the form $(x_1, x_2, \ldots)$ with finite support (i.e., finitely many non-zero elements) and let $f$ be the function that shifts an entire sequence one that shifts the entire sequence one coordinate to the right. Then $f$ is injective but not surjective. – Robert Shore Mar 25 '19 at 5:09