What does this notation mean? How can an isomorphism be a function? 
I'm extremely confused with this notation.
Could someone translate the second and third line of this example into something a little more declarative or understandable for me? 
On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?
Thanks so much
 A: It is a standard fact of linear algebra that any linear map $A: \>X\to Y$ between finite dimensional vector spaces with chosen bases $({\bf e}_k)_{1\leq k\leq n}$, resp. $({\bf f}_i)_{1\leq i\leq m}$, has a uniquely determined $(m\times n)$-matrix $[A]$  describing this map "computation wise". For each $(i,k)\in[m]\times[n]$ the  element $a_{ik}$ of this matrix is the $i^{\rm th}$ coordinate of the vector $A{\bf e}_k$:
$$A{\bf e}_k=\sum_{i=1}^m a_{ik}\,{\bf f}_i\ .$$
This general principle is now invoked when dealing with a differentiable function  $$f:\>(x,y)\mapsto(x^3,xy,y^2)\in{\mathbb R}^3\ .$$ This $f$  has a derivative $$df({\bf p}): \>T_{\bf p}\mapsto T_{f({\bf p})}\tag{1}$$ at the point ${\bf p}:=(2,3)$. 
This derivative is a map $(1)$, and therefore has a matrix $[df({\bf p})]\in M_{3,2}({\mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${\bf p}$:
$$[df({\bf p})]=\left[\matrix{{\partial f_1\over\partial x}&{\partial f_1\over\partial y}\cr
{\partial f_2\over\partial x}&{\partial f_2\over\partial y}\cr {\partial f_3\over\partial x}&{\partial f_3\over\partial y}\cr}\right]_{\bf p} =\left[\matrix{3x^2&0\cr
y&x\cr 0&2y\cr}\right]_{(2,3)}=\left[\matrix{12&0\cr 3&2\cr0&6\cr}\right]\ .$$
