Understanding Dual Transformations and reasoning behind definition In Linear Algebra working with Dual space and dual transformations I've come along this very basic definition of the dual transformations:

Suppose:
$T^*$ is a dual transformations from $W^*\to V^*$
   $T$ is a linear transformation from $V\to W$
   $u$ is a linear functional which belongs to $W^*$
   $v$ is a vector which belongs to $V$

   Then the following applies:
     $$ (T^*u)(v)  = u(Tv) $$ 

Why is the dual transformation defined this way? (I know this is a very problematic question, but please any intuition will be very helpful)
 A: In the special case of matrix algebra, this turns out to be fairly obvious.
In this setting, one usually writes vectors as $n \times 1$ matrices ("column vectors"), and linear functionals as $1 \times n$ matrices ("row vectors").
If we have a matrix $A$ with suitable dimensions, then "multiplication on the left" results in a linear transformation (call it $T$): i.e. $T(v) = Av$. The dual transformation is "multiplication on the right". That is, $T^*(u) = uA$. So your identity is merely
$$ (uA)v = u(Av) $$
A: Given a map $T: V \to W$ where $V, W$ are $K$ vector spaces.  We define the transpose or (dual transformation) $T^* :  W^* \to V^*$ of $T$ by $T^*(f) = f \circ T$.   I think it's defined this way, because it's the most natural way to associate to each linear map a new map on the corresponding dual spaces.  
There's a reason this assignment is called the transpose. If $V$ is finite dimensional, any basis $\beta = \{v_1, \dots, v_n\}$ of $V$, gives rise to $n$ linear functionals $v^*_i : V \to K$ defined by $v^*_i(w) = a_i$ where $a_i$ is the $i$'th component of $[w]_B$. The collection $B^* = \{v^*_i : 1 \le i \le n\}$ is linearly independent and spans $V^*$, and therefore forms a basis which we call the dual basis of $\beta$.  Fix ordered bases $\beta$ and $\gamma$ of $V$ and W, respectively.  Assuming $W$ is finite dimensional, if $T: V \to W$ is linear map, we have a matrix representation of $T$ denoted $[T]_\beta^ \gamma$.  It's fairly straightforward to prove that 
$$[T^*]_{\gamma^*} ^{\beta^*} = ([T]_\beta^ \gamma)^t$$
where by $( \cdots )^t$ I mean the transpose of a matrix.  
Some properties of taking the dual:
The assignment $V \mapsto V^*, T \mapsto T^*$ is what's called a contravariant functor (i.e. a functor that reverses composition and morphisms (linear maps)) on $\mathbf{Vect}_K$ the category of $K$ vector spaces.
If $V$ and $W$ are finite dimensional, $T \mapsto T^*$ is an injective linear map from $\operatorname{Hom}_K(V, W)$ to $\operatorname{Hom}_K(W^*, V^*)$ and hence an isomorphism since both spaces have dimesion $\dim V \dim W$.
