If linear map onto itself is bijective given that it is injective

Question

If given $l \in L(V)$ is injective then can we say that it is bijective too?
My reasoning:
by rank nullity theorem $N(l) = \{0\}$, implies $dim(R(l)) = dim(V)$, so $l$ is also surjective

Answer 1

It's a provable theorem that a linear map $T \in \mathcal{L}(V)$ is injective if and only if it is surjective, provided that $V$ is finite-dimensional. The argument is almost exactly as you suggested:

\begin{align*} \text{$T$ is injective } & \iff \text{Null}(T) = \{0\} \\ & \iff \dim V = \dim \text{Null}(T) + \dim \text{Range}(T) = \dim \text{Range}(T) \\ & \iff \text{Range}(T) = V. \end{align*} The third line follows from the second since $\text{Range}(T)$ is a subspace of $V$, so if $\dim V = \dim \text{Range}(T)$, we have $V = \text{Range}(T)$.

This result doesn't generalize to the infinite case. Notice that we used the rank-nullity theorem in the second line, which requires that $\dim V$ be finite.