Equivalence of surjectivity and injectivity for linear operators on finite dimensional vector spaces I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:V\rightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by
Surjectivity of $T \leftrightarrow \forall w \in V, \exists v \in V$ s.t. $ Tv = w.$
Let $v = v^ie_i$ for some basis $\{e_i\}$ of $V$.
$w = v^i(T e_i) = v^ie'_i$.
Surjectivity of $T$ now implies that the $\{ e'_i\}$ are another (linearly independent) set of basis vectors.
Linear independence of $\{e'_i\}$ implies that $i\neq j \rightarrow e_i'-e'_j \neq 0$ or $ e_i'-e'_j = 0 \rightarrow i = j$ or $Te_i = Te_j \rightarrow e_i = e_j \leftrightarrow T$ is injective.
Firstly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?
 A: Do you have the Rank-Nullity theorem yet?  
If so it's an easy application. 
If $n=\operatorname {dim}V$, then $\operatorname {rank}T+\operatorname {nullity}T=n$, so $T$ has full rank (is surjective)  precisely when the nullity is $0$  ($T$ is injective).
A: No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $\left\{e_n\right\}_{n\in{\mathbb N}}$, then
$$Te_i=e_{i+1}\ \forall i\in{\mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1}\ \forall i\ge 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces. 
A: Assume T is surjective. Let $\{e_n\}_{n=1}^N$ be a basis for V. By assumption $\{Te_n\}_{n=1}^N$ covers V. Then $\{Te_n\}_{n=1}^N$ is a basis for V since it contains the same number of elements as $\{e_n\}_{n=1}^N$ which is a basis (hence is non-redundant). Then for $v_1,v_2\in V,\ Tv_1=Tv_2\Rightarrow v_1=v_2.$
Assume T is injective. Then $\{Te_n\}_{n=1}^N$ is linearly independent. Therefore it  contains the same number of elements as a basis ($\{e_n\}_{n=1}^N$), hence $\{Te_n\}_{n=1}^N$ covers V. 
A: Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $\left\{v_1,\ldots,v_m\right\}$ be a basis of $ker(T)$ and complete to a basis $\left\{v_1,\ldots,v_n\right\}$ of $V$. Here $n\ge m\ge 1$.
Surjectivity implies that $\left\{Tv_{m+1},\ldots,Tv_n\right\}$ is a basis of $V$. Indeed any $w\in V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+\ldots+a_nv_n$, so $w=a_1Tv_1+\ldots+a_nTv_n=a_{m+1}Tv_{m+1}+\ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$. 
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$. 
Injectivity ===> Surjectivity
The argument is similar. For some basis $\left\{v_1,\ldots,v_n\right\}$ consider the images $\left\{Tv_1,\ldots,Tv_n\right\}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity. 
