I am trying to prove that $\phi: L(V,V) \to L(V^*, V^*)$ defined as $\phi(T) = T^t$ forms an isomorphism between $L(V,V)$ and $L(V^*, V^*)$.
notation:
$L(V,W)$ is the space of linear transformations from $V$ to $W$ (finite dimension vector spaces), $V^*$ is the dual space of $V$, and $T^t$ is the transpose of $T$
I can see that the simplest procedure would be to construct the inverse, however to do this I must use the fact that $(T^t)^t = T$ which I'm not quite sure how to do.
I'd instead like to show $\phi$ is bijective. Preferably, without resorting to the matrix representation.
Injective: $T^t = S^t$ iff $T = S$ for $S,T \in L(V,V)$
$T^t = S^t$ implies
$$f(Sv) = (S^t f)(v) = (T^t f)(v) = f(Tv)$$ For every $f \in V^*$ and $v \in V$
It feels intuitive that this would imply $S=T$, but proving that has not been my friend.
(And similar statement exists for $T^t \ne S^t$)
You proved that $f(Sv-Tv)=0$, for each $f\in V^*$. If $Sv\neq Tv$, then $Sv-Tv\neq0$ and you can always find a $f\in V^*$ such that $f(Sv-Tv)\neq0$. Therefore, $Sv=Tv$.
So, $\phi$ is injective. Since we are dealing with finite-dimensional spaces here, it must be surjective too.