What exactly is a matrix transformation? I'm confused on the notation for $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ and its relationship to linear transformations. We say that $\mathbb{R}^{m}$ is just the space of column vectors since there are n columns, right? 
If so, then what do we mean when we say T: $\mathbb{R}^{m} \mapsto \mathbb{R}^{n}$ is equivalent to $T_A(X) = AX$? I watched a video on linear algebra (essence of Linear Algebra) that (only if I'm remembering correctly) that all matrix multiplication was is the change of the standard basis vectors such as $\hat{\imath} \space \hat{\jmath}$ landed on different spots in the same dimension. 
What is exactly going on here? Can you have linear transformations that don't depend on the number of rows or columns to be the same? Geometrically, what is going on? It seems like I have no idea what is going on in a matrix at all. 

Well. To avoid confusion, let us say a linear transformation is mapping $T:V\rightarrow W$ between two abstract vector spaces over $\mathbb{C}$, $V$ and $W$, with $\dim V =n$ and $\dim W= m$. 
We say $T$ is a linear transformation if it satisfies the following properties


*

*$T(x+y) = T(x)+T(y) \ \ \forall \ \ x, y \in V$

*$T(\lambda x) = \lambda T(x) \ \ \forall \  \ \lambda \in \mathbb{C}$. 


The definition is very general and has nothing to do with matrices (not yet). 
Now, suppose we choose a basis for $V$, say $B_1=\{v_1, \ldots, v_n\}$, and a basis for $W$, say $B_2=\{w_1, w_2, \ldots, w_m\}$. Hence each vectors $v \in V$ will have a unique representation with respect to the basis $B_1$, i.e. we can write
\begin{align}
v = \alpha_1v_1+\alpha_2v_2+\ldots + \alpha_nv_n.
\end{align}
If we collect the coefficient into an array say $(\alpha_1, \ldots, \alpha_n)$, then this array uniquely determines $v$ since the representation is unique. In this process of identifying the abstract vector $v$ with its array of coefficients, we have essentially identified $V$ with $\mathbb{R}^n$. Likewise, we can identify $W$ with $\mathbb{R}^m$. 
Well. Once we have an identification, we would also like to see how $T$ works under this new identification, i.e. how would the coefficients transform with respect to the linear map $T$? Observe by linearity of $T$, we have
\begin{align}
T(v) = \alpha_1T(v_1) + \ldots +\alpha_nT(v_n). \ \ (\ast)
\end{align}
Since $T(v_i) \in W$, then they could be represented in terms of the basis $B_2$, i.e.
\begin{align}
T(v_j) = \beta_{1j}w_1 + \beta_{2j}w_2 + \ldots + \beta_{mj}w_m = \sum^m_{i=1} \beta_{ij}w_i. \ \ \ (\ast\ast)
\end{align}
Substitue $(\ast\ast)$ back into $(\ast)$ yields
\begin{align}
T(v) = \sum^n_{j=1}\alpha_j\sum^m_{i=1}\beta_{ij} w_i = \sum^m_{i=1}\left( \sum^n_{j=1} \beta_{ij}\alpha_j\right)w_i
\end{align}
which means the coefficients transform as follow
\begin{align}
(\alpha_1, \ldots, \alpha_n) \in \mathbb{R}^n \longmapsto  \left(\sum^n_{j=1} \beta_{1j}\alpha_j, \sum^n_{j=1} \beta_{2j}\alpha_j, \ldots, \sum^n_{j=1} \beta_{mj}\alpha_j \right) \in \mathbb{R}^m
\end{align}
One should note that the above transformation of the coefficients is actually given by
\begin{align}
\begin{pmatrix}
\beta_{11} & \beta_{12} & \ldots & \beta_{1n}\\
\beta_{21} & \ddots & \ldots & \vdots\\
\vdots & \ldots & \ddots & \vdots\\
\beta_{m1} & \ldots & \ldots & \beta_{mn}
\end{pmatrix}
\begin{pmatrix}
\alpha_1\\
\alpha_2\\
\vdots\\
\alpha_n
\end{pmatrix}.
\end{align}
Hence we could identify $T$ with the transformation of the coefficients via multiplication by a matrix $(b_{ij})$. In general, textbooks like to denote $(b_{ij})$ by $[T]_{B_1, B_2}$.
Example Let us consider the following example to conclude our discussion. Consider $V = P_2$, the space of polynomials with degree at most $2$, and $W = P_1$, the space of polynomials with degree at most $1$. Let us define the linear $T$ given by
\begin{align}
T(a_0+a_1 x+a_2x^2):= \frac{d}{dx}(a_0+a_1 x+a_2x^2)= a_1+2a_2 x.
\end{align}
It's clear that $T$ is a linear map from $P_2$ to $P_1$. 
Let us choose a basis  $B_1=\{1, x, x^2\}$ for $V$ and a basis $B_2=\{1, x\}$ for $W$, then we see
\begin{align}
\left[ \frac{d}{dx}\right]_{B_1, B_2} =
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 2\\
\end{pmatrix}.
\end{align}
Edit: A fast way to determine the matrix representation of a linear transformation is to use the following identity
\begin{align}
[T]_{B_1, B_2} = ( [T(v_1)]_{B_2}, [T(v_2)]_{B_2},\ldots, [T(v_n)]_{B_2}).
\end{align}
For the above example, we see that
\begin{align}
T(1) = 0= 0\cdot 1 + 0\cdot x,  \ \ T(x) = 1 = 1\cdot 1 + 0\cdot x \ \ \text{ and } \ \ T(x^2) = 2x = 0\cdot 1 + 2\cdot x
\end{align}
where $\{1, x, x^2\}$ is the basis for $V$ which mean
\begin{align}
[T(1)]_{B_2} = 
\begin{pmatrix}
0 \\
0\\
\end{pmatrix}, 
\ \ 
[T(x)]_{B_2} = 
\begin{pmatrix}
1 \\
0\\
\end{pmatrix}
\ \ \text{ and } \ \
[T(x^2)]_{B_2} = 
\begin{pmatrix}
0 \\
2\\
\end{pmatrix}.
\end{align}
So together we have
\begin{align}
[T]_{B_1, B_2} =( [T(v_1)]_{B_2}, [T(v_2)]_{B_2},\ldots, [T(v_n)]_{B_2})= \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 2\\
\end{pmatrix}.
\end{align}
