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?
Best Answer
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.