Why is an ${m \times n}$ matrix a transformation from $\mathbb R^n$ to $\mathbb R^m$? Probably basic, but I just don't get the concept.
 A: Taking a matrix $A \in R^{m \times n}$, a vector $v \in R^n$, how many entries does $A\cdot v$ have?
A: A $m \times n$ matrix $A$ is not a transformation!
Rather, given basis of $\mathbb{R}^n$ and $\mathbb{R}^m$, it represents a transformation $u$.  
eg say you have a basis $v_1, \dots, v_n$ of $\mathbb{R}^n$ and a basis $w_1,\dots, w_m$ of $\mathbb{R}^m$. Let $u : \mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear function.
Any vector $v \in \mathbb{R}^n$ can be written uniquely 
$v = \sum_{i=1}^n \lambda_i \cdot v_i$ for some scalars $(\lambda_i)_{1\leqslant i \leqslant n} \subset \mathbb{R}$. One can see that $u(v) = \sum_{i=1}^n \lambda_i \cdot u(v_i)$.
Hence, $u$ is determined by the way it acts on the basis  of $\mathbb{R}^n$.
In other words, a compact way of describing $u$ is to describe only the vectors $u(v_1),\dots,u(v_n)$. We will now describe these vectors in the basis $w_1,\dots,w_m$ :
Take any $i$, there are scalars $\alpha_{i,1},\dots,\alpha_{i,m}$ such that 
$$u(v_i) = \sum_{j=1}^m \alpha_{i,j} \cdot w_j$$
So, given these two basis, $u$ is fully described by the matrix
$$A := \begin{pmatrix} \alpha_{1,1} & \dots & \alpha_{n,1}\\
\vdots & \ddots & \vdots \\
\alpha_{1,m} & \dots & \alpha_{n,m} \end{pmatrix}$$
A: This is not so much an answer as an observation. We normally write functions on the left, as in applying a linear transformation $T$ to a vector $v \in \mathbb{R}^n.$ So, when we determine a matrix representation $A$ of $T,$ say the standard matrix, we multiply $v$ on the left by $A,$ treating $v$ as a $1$ column matrix, so $A$ will be $m \times n$ if we wish to obtain a vector in $\mathbb{R}^m.$
When I first learned linear algebra years ago, there was a fashion among some algebraists to write functions and operators on the right, so that the composition of functions $f$ and $g$ would be written $f \circ g,$ the natural left to right order. In this situation, the value of $T$ at $v$ would be $vT.$ Now, the matrix multiplication by the standard matrix $A$ of $T$ would be on the right, so we have $vA,$ treating $v$ as $1 \times n$ matrix. In this setup, $A$ would indeed be an $n \times m$ matrix, probably what people would expect. 
Although I have read some algebra where authors used "function on the right" notation, my original exposure to linear algebra was the only time I actually used it. It took a bit of concentration to deal with the traditional "function on the left" when I met it in later encounters with linear algebra. By now, it is second nature. The right-hand notation has its benefits, but the tradition of left-hand notation is difficult to oppose.
