Let $V$ be a finite-dimensional vector space over the field $F$. Show that $T \to T^t$ is an isomorphism of $L(V, V)$ onto $L(V^*, V^*)$.
Note: For $T\in L(V, W) $ , the dual map (or transpose) $T^t\in L(W^{\star}, V^{\star})$ is defined by $$(T^tg) (\alpha) =g(T\alpha) $$ for every $g\in W^{\star}$ and $\alpha\in V$
We already known $L(V,V)\cong L(V^*,V^*)$, since $\mathrm{dim}(L(V,V))$ $=(\mathrm{dim}(V))^2$ $= (\mathrm{dim}(V^*))^2$ $= \mathrm{dim}(L(V^*,V^*))$. In this problem, we are being asked to show $f:L(V,V)\to L(V^*,V^*)$ defined by $f(T)=T^t$ is isomorphism.
My attempt: Let $T,L\in L(V,V)$ and $c\in F$. Then $f(c\cdot T+L)$ $=(c\cdot T+L)^t$. So $[(c\cdot T+L)^t](g)$ $=g\circ (c\cdot T+L)$ $=c\cdot (g\circ T)+g\circ L$ $=c\cdot T^t(g)+L^t(g)$ $=(c\cdot T^t+L^t)(g)$, $\forall g\in V^*$. Thus $(c\cdot T+L)^t$ $=c\cdot T^t+L^t$ $=c\cdot f(T)+f(L)$. Hence $f$ is an linear map.
Now we show $f$ is bijective. Since $\mathrm{dim}(L(V,V))$ $= \mathrm{dim}(L(V^*,V^*))$, it’s suffice to show either $f$ is injective or surjective, by theorem 9 section 3.2. If $f(T)=f(L)$, for some $T,L\in L(V,V)$. Then $T^t=L^t$, i.e. $T^t(g)$ $=L^t(g)$ $=g\circ T$ $=g\circ L$, $\forall g\in V^*$. If we can find an injective map $g\in V^*$$=L(V,F)$, then we’re done. I tried following linear functional on $V$, $g(\alpha_j)=1_F$, where $\{\alpha_1,…,\alpha_n\}$ is basis of $V$. But that isn’t working. Another way to show injective, $N_f=\{0\}$. Let $T\in N_f$. Then $f(T)$ $=T^t$ $=0$, i.e. $T^t(g)$ $=0(g)$ $=g\circ T$ $=0$, $\forall g\in V^*$. Again I’m running into same problem. We have to find an injective map $g\in L(V,F)$ so that $g(T(\alpha))=0$$\implies$$T(\alpha)=0$, $\forall \alpha \in V$. I think showing surjectivity is even more difficult than injective. We are literally working in inception of inception of maps. How to progress from here?
This is easy since after identifying $V^{**} = V$, we have $(T^t)^t = T$. The identification $V \approx V^{**}$ identifies $v \in V$ with the map $\omega \mapsto \langle v, \omega \rangle_{V, V^*}$. It may seem like this identification is only an inclusion, but surjectivity follows since $V$ and $V^{**}$ are both $n$-dimensional.
Edit: $(T^t)^t = T$ because for $v' \in V^*$ and $v \in V^{**} = V$, $$\langle T^t v', v \rangle = \langle v, T^t v'\rangle = \langle Tv, v' \rangle = \langle v', Tv\rangle.$$