Transpose of an invertible linear transformation.. 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.
 A: 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. 
A: Your proof looks fine to me, but it's possible to give a simpler argument. 
Suppose $T\colon V\to W$ is invertible with inverse $S\colon W\to V$. I claim that $T^t\colon W^*\to V^*$ is invertible with inverse $S^t\colon V^*\to W^*$. Indeed, for any $f\in V^*$, $T^t(S^t(f)) = T^t(fS) = fST = f$ since $ST = \text{id}_V$, and for any $G\in W^*$, $S^t(T^t(g)) = S^t(gT) = gTS = g$, since $TS = \text{id}_W$.
A: I think there is a point that must be clarified: does the dimension of $V$ equals the dimension of $W$, or not? If the answer is not then you have to distinguish between left and right inverse.
Let's consider the matrix case for a moment. In this setting the question is: is $T$ a square matrix?
If $T$ is a square matrix then the following are equivalent


*

*$T$ is injective 

*$T$ is surjective 

*$T$ is bijective 

*$\det(T)\neq 0$

*$T$ admits a left inverse, i.e. $\exists \; L$ such that $LT=\mathbb{1}$

*$T$ admits a right inverse, i.e. $\exists \; R$ such that $TR=\mathbb{1}$


On the other hand, if $T$ is not a square matrix, then $T$ could have a right inverse and not a left one or viceversa. If I'm not wrong we can say


*

*$T$ is injective $\iff$ $T$ admits a left inverse

*$T$ is surjective $\iff$ $T$ admits a right inverse


All of this preamble to say that, in your argument, you are assuming that $T$ has a right inverse, precisely when you say 

T is invertible so we have a $T^{-1}$ such that $ \;(g_1-g_2)TT^{-1} = g_1-g_2 $.

Therefore I believe your argument is correct in the case $\dim(V)=\dim(W)$, but it is wrong in the general case. I think in the general setting what holds is the following
$$T \text{ is injective} \iff T^t \text{ is surjective} $$
