Why the transpose of $A$ is not looking at the transformation $A$ in the dual space Most of the time people are saying that in order to have a better understanding about what the transpose really represents is this:

We have a linear transformation $T : V \to V$ with $V$ a finite vector space.  Then the tranpose is a linear application in the dual space, namely, $T^* : V^* \to V^*$ such that $T^*(\phi)(v) = \phi(Tv)$.

I understand what all of this means, but what is the motivation behind this definition of the transpose? 
For me, if we have a linear map $T : V \to V$, then a more natural thing to look at is this:  what is the transformation $T$ in the dual space $V^*$ ?
 In this case this is simply the linear application $T^* : V^* \to V^*$ such that : $T^*(\langle y, . \rangle) = \langle Ty, . \rangle$. For me this is a natural transformation to look at, namely, the application $T$ acting on the dual space. 
Yet I don't understand the motivation behind the transpose. Why this definition, and not another? 
Thank you! 
 A: The transpose map $T^*\colon W^*\to V^*$ can be defined for every linear map $T\colon V\to W$, where the vector spaces need not be endowed with an inner product and could be infinite dimensional as well.
The case of $T\colon V\to V$ where $V$ is a finite dimensional inner product space (over the reals, for simplicity) is a bit different, but not really too much.
The inner product allows you to define an isomorphism $\Phi\colon V\to V^*$, via the mapping $v\mapsto\Phi(v)=\hat{v}$, where $\hat{v}(w)=\langle v,w\rangle$.
This mapping is linear and injective, because $\langle v,w\rangle=0$ for all $w$ implies $v=0$. Hence, due to $V$ being finite dimensional, $\Phi$ is an isomorphism.
Now consider the map $T'\colon V\to V$ defined by
$$
T'=\Phi^{-1}\circ T^*\circ\Phi
$$
where $T^*\colon V^*\to V^*$ is defined by $T^*(\varphi)=\varphi\circ T$.
Consider now $v\in V$; then $\Phi\circ T'(v)=\Phi(T'(v))=\widehat{T'(v)}$ is the map such that
$$
\widehat{T'(v)}(w)=\langle T'(v),w\rangle
$$
by definition. On the other hand, $T^*\circ\Phi(v)=T^*(\hat{v})=\hat{v}\circ T$ is the map such that
$$
\hat{v}\circ T(w)=\hat{v}(T(w))=\langle v,T(w)\rangle
$$
Thus $T'$ is exactly the unique linear map $T'\colon V\to V$ such that, for all $v,w\in V$,
$$
\langle T'(v),w\rangle=\langle v,T(w)\rangle
$$
and the argument above is precisely the proof that this $T'$ exists.

What if $T\colon V\to W$ and both spaces are finite dimensional? To every basis $\mathscr{B}$ of a vector space, there corresponds a unique dual basis in the dual space, $\mathscr{B}^*$. If $\mathscr{B}$ is a basis of $V$ and $\mathscr{D}$ is a basis of $W$, we can compute the matrix $A$ of $T$ with respect to these basis. Then the matrix of $T^*\colon W^*\to V^*$ with respect to the dual bases is precisely the transpose matrix of $A$.
