Prove the surjectivity of this injective linear map

Question

I am working on the following problem.

Let $g : V\to V$ be linear and injective, where $V$ is a vector space over the field K.
Prove that, if $V$ is finite-dimensional, then $g$ is surjective.

In an attempt to solve the problem, I have done the following.

I first chose a basis $\{e_1, \ldots, e_n\}$ for $V$ (assuming that $\dim(V) = n$) and considered the images under g of $e_1, \ldots, e_n.$

I then attempted to prove the required statement by contradiction.
I assumed that $g$ is not surjective.
Then $\exists \space v_d\in V$ s.t. (such that), $\forall \space v_c \in V \space \forall \space \gamma_1, \ldots, \gamma_n \in K$, we have that $\gamma_1 g(e_1) + \ldots + \gamma_n g(e_n) \not= v_d.$ $(*)$

My next thought was that, if I could show that the vectors $g(e_1), \ldots, g(e_n)$ formed a basis of $V$, then $(*)$ would imply that no linear combination of these vectors was equal to $v_d$, contradicting the definition of a basis.
Consequently, I managed to show that $g(e_1), \ldots, g(e_n)$ are linearly independent.
However, I am now struggling to show that $g(e_1), \ldots, g(e_n)$ span $V$.
I would appreciate any hints which could help me progress.

NB:
Firstly, the question tells me not to use the Rank-Nullity Theorem.
Secondly, I am aware that, since $g$ is injective, $\ker(g) = \{0\}.$

Answer 1

Method 1

$Rank(g)+Nullity(g)=dim(V)$ is the well known equality. Now, $g$ is injective so $Nullity(g)=0$ hence $Rank(g)=dim(V)$. But $Range(g)\subset V$. Hence $Range(g)=V$ which shows $g$ is surjective.

Method 2

We will show that $g$ maps a basis of $V$ to a basis of $V$. Let dim(V)=n. Let $\{e_1,e_2,...,e_n\}$ be any basis of $V$. We will show that $\{g(e_1),g(e_2),...,g(e_n)\}$ is a basis of $V$.

It is sufficient to show linear independence.

Let there exist scalars $\alpha_i$ such that $\sum_{i=1}^n\alpha_ig(e_i)=0$.

This means, $g(\sum_{i=1}^n \alpha_ie_i)=0$ but $g$ is injective, so $\sum_{i=1}^n\alpha_ie_i=0$. But $\{e_1,...,e_n\}$ is a basis, hence we must have $\alpha_i=0\forall i$. This proves linear independence of $\{g(e_1),...,g(e_n)\}$. Since this is a set of vectors (linearly independent) with same cardinality as $dim(V)$ we conclude this set is a basis of $V$.

Hence given any element $v\in V$ there exist scalars $b_i$ such that $\sum_{i=1}^nb_ig(e_i)=v$ i.e. $g(\sum_{i=1}^n b_ie_i)=v$ which shows surjectivity of $g$.