I am trying to prove that suppose that a linear transformation $T$ is invertible, then its transpose $T^t$ is also invertible.
Is the following proof correct?
Proof:
Let $T$ be an invertible linear transformation that maps elements from the vector space $V$ to $W$. Let $V^\ast$ and $W^\ast$ be the dual spaces respectively.
Suppose that $T^t(g_1)=T^t(g_2)$ where $g_i$ is an element in $W^\ast$. By definition, $T^t(g)=gT$. Therefore, $g_1T=g_2T$. Adding the additive inverse of $g_2T$ to both sides of the equation, we get $g_1T-g_2T=0$ and therefore $(g_1-g_2)T=0$.
Since $T$ is invertible, $T^{-1}$ exist. Applying both sides of $(g_1-g_2)T=0$ to $T^{-1}$, we get $(g_1-g_2)TT^{-1}=0T^{-1}$ and $(g_1-g_2)I=0$. Hence $g_1-g_2=0$.
Finally, this means that $T^t(g_1)=T^t(g_2)$ implies $g_1=g_2$, proving that $T^t$ is injective.
To show that $T^t$ is surjective, we need to show that for all $f$ in $V^\ast$, there exist some $g$ in $W^\ast$ such that $T^t(g)=f$, or $gT=f$. Since $T$ is invertible, given any such $f$, a $g$ can be found and is given by $fT^{-1}$. Hence, $T^t$ is surjective.
Since $T^t$ is bijective, it is invertible.
Best Answer
I think we can prove a little stronger result that
$$ T \text{ is surjective } \implies T^{t} \text{ is injective}. $$
But let's do it by putting all the definitions first.
Let $ V $ and $ W $ be linear spaces over a field $ F $.
$T : V \mapsto W $ be an isomorphism i.e a linear transformation that is injective and surjective. Of course, as a bijective map, there is a map $ T^{-1} : W \mapsto V $ and one can prove that this map is itself a linear transformation.
Let $ l : V \mapsto F $ be a "linear functional" i.e. a linear transformation from $ V $ to the one-dimensional vector space $ F $.
The set $ V^{*}= \left \{ l| l : V \mapsto F \text{ is a linear functional } \right \} $ forms a vector space over $ F $, with addition and scalar multiplication defined in the obvious way. We call it the dual of $ V $. Similarly, let $ W^{*} $ denote the dual of $ W $.
If $ l \in W^{*} $, then the composition $ l \circ T : V \mapsto F $ is a linear functional over $ V$, and so we get an assignment, $ T^{t} $, which sends $ l \in W^{*} $ to $ m = l \circ T $ in $ V^{*} $. We can prove that $ T^{t} : W ^{ * } \mapsto V ^ { * } $ is a linear transformation, such that, for $ v \in V $, we have
$$ (T^{t}( l)) ( v ) = l ( T ( v ) ) = m(v). $$
We now prove the following:
(1) If $ T $ is surjective, then $ T^{t} $ is injective.
(2) If $ T $ is invertible, then $ T^{t} $ is surjective.
$ \textit{Proof:} $ We prove (1) first. Suppose that $ T^{t}(g_{1}) = T^{t} ( g_{2} ) $ where $ g_{1}, g_{2} \in W^{*}$. Then, $ g_{1}T = T^{t}(g_{1} ) = T^{t}(g_{2}) = g_{2}T $. Thus, for every $ v \in V $, we have that
$$ g_{1}(T(v)) = g_{2}(T(v)). $$
As $ T $ is surjective, every $ w \in W $ can be written as $ T(v) $ for some $ v \in V $, and so, the last equality says that $ g_{1} $ and $ g_{2} $ agree on every $ w \in W $, and so $ g_{1} = g_{2} $.
To prove (2), let $ m \in V^{*} $, then $ m \circ T^{-1} \in W ^ { * } $, and we have
$$ T^{t} ( m T^{ - 1} ) = (m T^{-1} ) T = m( T^{-1} T) = m ( I ) = m. $$
$ \textit{ Remark : } $ The only thing that I don't like about your proof is when you "apply" the linear functional $ ( g_{1} - g_{2})T $ to $ T^{t} $. You are composing two functions. Don't say apply. The functional $ ( g_{1} - g_{2} ) T $ can only be applied to vectors $ v \in V $. Otherwise, the proof seems fine.