Let $L:V \to W$ and $V$ is finite-dimensional. Show that if $\dim\ker(L)={0}$, then $V\cong L(V)$ 
Let $L:V \to W$ and $V$ is finite-dimensional. Show that if $\dim \ker(L)={0}$, then $V\cong L(V)$.

According to the Rank-nullity theorem $$r=0+\dim L(V),$$
where $r$ is the dimension of $V$, the domain $L$. It's clear that $V$ and the range of $L$, $L(V)$, have the same dimension. If we can show that $L(V)$ is a subspace of $V$, we can then claim that they are equivalent as the implication requires. 
A trivial nullspace means $L$ is injective, and among other things this means that if $E=\{e_1,...,e_2\}$ is a basis for $V$, then the set $L(E)$ is also independent. I've been trying to work this through via bases and the facts stated above. Should I look into isomorphisms ( or something else ) or do I have enough information here as it is to write a proof? 
 A: If I got it right, you want to prove that $L(V)$ is a linear subspace of $W$. If it is the case, you can proceed as next.
Proposition
Given two finite dimensional vector spaces $V$ and $W$ and a linear transformation $L:V\rightarrow W$, then $L(V)\leq W$.
Proof
Since $L(V)\subseteq W$, we have to prove that $aw_{1} + w_{2}$ belongs to $L(V)$ whenever $w_{1},w_{2}\in L(V)$ and $a\in\textbf{F}$.
Indeed, if $w_{1}\in L(V)$, there is a vector $v_{1}\in V$ such that $T(v_{1}) = w_{1}$. Similarly, if $w_{2}\in L(V)$, there exists $v_{2}\in V$ such that $T(v_{2}) = w_{2}$. Since $av_{1} + v_{2}\in V$, we do also have that $T(av_{1} + v_{2})\in L(V)$.
Due to the linearity of $T$, one has that
\begin{align*}
av_{1} + v_{2} \Longrightarrow T(av_{1}+v_{2}) \in L(V) \Longrightarrow aT(v_{1}) + T(v_{2}) = aw_{1} + w_{2}\in L(V)
\end{align*}
and we are done.
Hopefully this helps.
A: What you are trying to prove is wrong as is : if you have no idea what the elements of $V$ and those of $W $are, and there is no reasons for elements of $L(V)$, which are elements of $W$ to even be in $V$.
For example $V$ could be the space of polynoms whose degree is $\leq n$ $$ V =\mathbb{R}_n[X]$$
$W$ could be a space of tuples
$$W = \mathbb{R}^n$$
there are functions from $V$ to $W$ such as polynom evaluation on a set on $n$ distinct points $(a_1, \ldots, a_n)$ :
$$ L : P \in \mathbb{R}_n[X] \mapsto (P(a_1), \ldots, P(a_n)) \in\mathbb{R}^n $$
which verifies $ ker(L) = 0$ (by Lagrange’s theorem)
But elements of $L(V)$ are not elements of $V$ since they are tuples and not polynomials !
It is nevertheless true that (as pointed out in comment above) $ker(L) = 0$ implies that $L^{|L(V)}$ is bijective, and therefore that $V$ and $L(V)$ are isomorphic (but not equal).
