Represent the transformation with respect to the standard basis 
Consider a linear transformation from $R^2$ to $R^2$ defined via:
$$\left(\begin{matrix} 1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 1\end{matrix} \right)$$ and $$\left(\begin{matrix} -1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 2\end{matrix} \right)$$
Represent the transformation with respect to the standard basis

It has been some time I have tried to understand this problem, and I have seen solutions to problems similar to this and I still do not understand it. Those solutions talk about images of $e_1$ and $e_2$ (which are the standard vectors of the standard basis), but I don't know what the heck they are talking about.
 A: We can represent a linear map: $F:\mathbb{R}^{2} \to \mathbb{R}^{2}$ using a matrix $\mathbf{A}\in\mathbb{R}^{2\times2}$, generically, we can write:
$$\mathbf{A}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$
Applying this to the vector $\begin{pmatrix}1 & 3\end{pmatrix}^{T}$, we have:
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}1 \\ 3\end{pmatrix}=\begin{pmatrix}a+3b \\ c + 3d\end{pmatrix} = \begin{pmatrix}3 \\ 1\end{pmatrix}$$
Similarly, applying it to the vector $\begin{pmatrix}-1 & 3\end{pmatrix}^{T}$, we have:
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}-1 \\ 3\end{pmatrix}=\begin{pmatrix}-a+3b \\ -c+3d\end{pmatrix} = \begin{pmatrix}3 \\ 2\end{pmatrix}$$
We therefore have a system of 4 linear simultaneous equations with 4 variables $a, b, c$ and $d$:
$$\begin{align*}a+3b &= 3 \\ c+3d &= 1 \\ 3b-a &= 3 \\ 3d - c &= 2\end{align*}$$
Solving these we get:
$$a=0,\quad b=1,\quad c = -\frac{1}{2},\quad d=\frac{1}{2}$$
We can thus write our transformation matrix describing the linear map, as:
$$\mathbf{A} = \frac{1}{2}\begin{pmatrix}0 & 2 \\ -1 & 1\end{pmatrix}$$
A: A vector can be decomposed regarding a basis, esp. the standard basis:
$$
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
=
x
\left(
\begin{matrix}
1 \\
0
\end{matrix}
\right)
+
y
\left(
\begin{matrix}
0 \\
1
\end{matrix}
\right)
= x e_1 + y e_2
$$
so a linear transformation $A$ acting on a vector can be calculated by $A$ acting on the base vectors.
$$
A
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
=
x\,A
\left(
\begin{matrix}
1 \\
0
\end{matrix}
\right)
+
y\,A
\left(
\begin{matrix}
0 \\
1
\end{matrix}
\right)
=
x \, A e_1 + y \, A e_2
$$
The matrix of $A$ can be seen as
$$
A = \left((Ae_1) (Ae_2)\right)
$$
So from
$$
A
\left(
\begin{matrix}
1 \\
3
\end{matrix}
\right)
= 1\,A e_1 + 3\, A e_2 =
\left(
\begin{matrix}
3 \\
1
\end{matrix}
\right)
\quad
A
\left(
\begin{matrix}
-1 \\
3
\end{matrix}
\right)
= -1\,A e_1 + 3\, A e_2 =
\left(
\begin{matrix}
3 \\
2
\end{matrix}
\right)
$$
one can take the sum and difference:
$$
6\, A e_2 =
\left(
\begin{matrix}
6 \\
3
\end{matrix}
\right)
\quad
2\,A e_1 =
\left(
\begin{matrix}
0 \\
-1
\end{matrix}
\right)
$$
This gives
$$
A = \left((Ae_1) (Ae_2)\right) =
\left(
\begin{matrix}
0 & 1 \\
-1/2 & 1/2
\end{matrix}
\right)
$$
