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?

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. 
