Matrix associated with a projection mapping From S.L Linear Algebra:

Find the matrix associated with the following linear maps. The vectors
  are written horizontally with a transpose sign for typographical
  reasons.
  (a) $F:\mathbb{R}^4 \rightarrow \mathbb{R}^2$ given by $F\left ((x_1, x_2, x_3, x_4)^T \right)=(x_1, x_2)^T$ (the projection)

Solution Attempt
I will use Theorem 2.1 from the book:

Let $L: K^n \rightarrow K^m$ be a linear map. Then there exists a
  unique matrix $A$ such that $L = L_A$.

Hence, for all $X$ we have $L(X)=AX$.
In this case, some $A$ is the matrix associated with linear map $F$ such that:
$$F(X^T)=AX^T$$
$$F\left ((x_1, x_2, x_3, x_4)^T \right)=A\left ((x_1, x_2, x_3, x_4)^T \right)$$
$$AX^T=A\left ((x_1, x_2, x_3, x_4)^T \right)=(x_1, x_2)^T$$
Solving for $A$:
$$A=\left ((x_1, x_2, x_3, x_4)^T \right)^{-1}(x_1, x_2)^T$$
Now this makes no sense, because $X$ must be non-degenerate square matrix in order to be invertible, but $X$ is just a vector with cardinality (dimension) $4$.
Perhaps I could invert $A$ (then I would have to show that $F$ is injective with trivial kernel), but I don't see any point of doing this.
Is $A$ truly a matrix associated with $F$? If not, how to find $A$ if it is a matrix that is truly associated with $F$?

P.S
I also found this in the book:

Let $F: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ be the projection, in
  other words the mapping such that $F(x_1, x_2, x_3) = (x_1, x_2)$.
  Then the matrix associated with $F$ is:
$$\begin{pmatrix} 1 & 0  & 0 \\   0&1   & 0  \end{pmatrix}$$

I don't know how was it exactly calculated, but the matrix above contains standard basis vectors only, which I believe must mean something.
Thank you!
 A: To begin with, the Vector Space of Matrices $\mathscr{M}_{mn}(\mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $\mathbb{R}^2$ can be extended so to be isomorphic to $\mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(\left (
\begin{matrix}
 x_1 \\ 
 x_2 \\
 x_3 \\
 x_4 
\end{matrix}
\right )
)=\left (
\begin{matrix}
 x_1 \\ 
 x_2 
\end{matrix}
\right )
 $$
ie.
$$ \left (
\begin{matrix}
 a_{11} & a_{12} & a_{13} & a_{14} \\ 
 a_{21} & a_{22} & a_{23} & a_{24}
\end{matrix}
\right )
.\left (
\begin{matrix}
 x_1 \\ 
 x_2 \\
 x_3 \\
 x_4 
\end{matrix}
\right )
=\left (
\begin{matrix}
 x_1 \\ 
 x_2 
\end{matrix}
\right )
 $$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ \begin{cases}
 a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \\ 
 a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
\end{cases} $$
ie.
$$ \begin{cases}
 (a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \\ 
 a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
\end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $\mathbb{R}^4$ is:
$$
\left (
\begin{matrix}
 1 & 0 & 0 & 0 \\ 
 0 & 1 & 0 & 0 
\end{matrix}
\right )$$
