Matrices as linear transformations I am reading a proof which claims: 

A matrix of $m\times n$ is a linear transformation from $m$
  vector-space to $n$ vector-space, And therefore, by the dimension theorem: $m = \dim\ker A + \dim\Im A$

Isn't it the opposite?  
For example: $A\in M_{4\times 2}(\mathbb{R})$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^4$.

Let us take an example:
$$ \begin{pmatrix} \color{blue}1& \color{red}2 & \color{green}3 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} \color{blue}{10} \\ \color{red}{20} \\ \color{green}{30}\end{pmatrix} = \begin{pmatrix}\color{blue}1 \cdot \color{blue}{10} + \color{red}2\cdot \color{red}{20} +\color{green}3 \cdot \color{green}{30} \\ 380\end{pmatrix} $$
but
$$ \begin{pmatrix} \color{blue}1& \color{red}2 & \color{green}3 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} \color{blue}{10} \\ \color{red}{20} \end{pmatrix} = \begin{pmatrix}\color{blue}1 \cdot \color{blue}{10} + \color{red}2\cdot \color{red}{20} +\color{green}3 \cdot \color{green}{ ? } \\ ?\end{pmatrix} $$
seems to be harder to compute. So, if $\mathbb{R}^{m\times n}$ is set the set of matrices with $m$ rows and $n$ columns then for $A \in \mathbb{R}^{m\times n}$ and $x \in \mathbb{R}^n$ we have $Ax \in \mathbb{R}^m$.
