Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$
(i) If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.
(ii) Give example showing (i) is false if $V$ is not finite dimensional.
So on (i), since $V$ is finite dimensional, then $V$ has a basis with finite cardinality and hence $\dim(V)=n$. Also, the following holds, $\dim(V)=\dim(\ker(\alpha) + \dim(\operatorname{im}(\alpha))$ Since $\alpha$ is injective, then the $\ker(\alpha)$ is $0$ which implies that the $\dim(\ker(\alpha)$ is $0$ which implies that $\dim(\operatorname{im}(\alpha))=\dim(V)$ and that shows that $\alpha$ is surjective. (Is that right?)
On the second one, (ii), I just have no idea. I mean shouldn't Axiom of Choice be able to work here?
Best Answer
The statements (i) and (ii) illustrate that finite-dimensional vector spaces behave in a way similar to finite sets, for a map $X \to X$ on a finite set is injective iff it is surjective. However, if $X$ is an infinite set there are maps that are injective but not surjective (e.g. $\mathbb N \to \mathbb N, x \mapsto x+1$) and vice versa (e.g. $\mathbb N \to \mathbb N, x \mapsto \max\{1,x-1\}$). You can turn this into a counterexample for the corresponding statement for vector spaces: let $V$ be the vector space freely generated by $\mathbb N$, i.e. the vector space of sequences $(a_1,a_2,\ldots) \in k^{\mathbb N}$ where $a_i =0$ for all but finitely many $i$ and consider the linear maps induced by the above maps, i.e. $$(a_1,a_2,\ldots) \mapsto (0,a_1,a_2,\ldots)$$ and $$(a_1,a_2,\ldots) \mapsto (a_1+a_2,a_3,a_4\ldots)$$ respectively. The first map is injective but not surjective, and vice versa for the second map.