Is the function linear? Given
$$ F(x) = \left(\begin{matrix} -x_2 \\ x_1+2x_2 \\ 3x_1 - 4x_2 \end{matrix} \right), x = \left(\begin{matrix} x_1 \\ x_2 \end{matrix} \right)$$
Prove whether $F$ is a linear function or not.
I've tried to prove it, but I'm not sure it's right:
$$F(x) = Mx \Rightarrow M = \left(\begin{matrix}
0 & -1 \\ 
1 & 2 \\
3 & -4 \end{matrix}\right)$$
Let there be $u,v \in \mathbb{R}^2$, then:
$$\begin{eqnarray*}
F(u+v) = M (u+v) = Mu+Mv = F(u)+F(v)
\end{eqnarray*}$$
and
$$F(\lambda{u}) = M\lambda{u} = {\lambda}Mu = {\lambda}F(u)$$
so the function $F$ is linear.
Using the matrix $M$ was not shown in any of the examples I've seen, I've just invented the technique, I think it's nicer (and almost certainly I've reinvented the wheel). What's the technical term for $M$?
I'm sure there is a missing step in arriving from the given function to the form $F(x)=Mx$, M looks like the coefficient vectors arranged column-wise.
Is the proof formally correct?
 A: The main thing, is that there is a matrix $M$, such that $F(x)=M\cdot x$ for all vectors $x$. In fact for a mapping $F$ between vector spaces with fixed bases, this is equivalent to be linear. Then, that multiplying by a matrix is indeed a linear mapping, is straightforward: depend on the nice properties of matrix multiplication, and you proved it formally correctly.
So, what should be emphasized is that indeed $F(x)=M\cdot x$ for all $x$, but this can be easily seen by expanding the product on the right hand side.
$M$ is called the matrix belonging to the linear map $F$, with respect to the standard bases.
Let $V$ and $W$ be vector spaces, with basis $\mathcal B:=(b_1,..,b_n)$ in $V$ and basis $\mathcal C:=(c_1,..,c_m)$ in $W$. If $F:V\to W$ is a linear mapping, then define its matrix wrt. bases $\mathcal B$ and $\mathcal C$ as:
$$[F]^{\mathcal B}_{\mathcal C} := \left[ [F(b_1)]_{\mathcal C}\, ... [F(b_n)]_{\mathcal C} \right] $$
where $[w]_{\mathcal C}$ denotes the culomn vector $\in \Bbb R^m$ of coordinates of $w$ in basis $\mathcal C$, i.e. 
$$[w]=\pmatrix{\alpha_1\\ \vdots \\ \alpha_m} \iff w=\alpha_1 c_1+..+\alpha_m c_m. $$
Then, by linearity one can check that for all $v\in V$ we have
$$[F]^{\mathcal B}_{\mathcal C}\cdot [v]_{\mathcal B} = [F(v)]_{\mathcal C}\ . $$
Prove it first for the basis vectors $v=b_i$.
A: You've shown that your map can be represented by a matrix and then shown that it must be linear based on the properties of matrix multiplication.  
In fact, most mathematicians would view the problem the other way round: matrices are introduced as a neat way of encoding a linear map: so while it is certainly true that any map that can be represented by a matrix $A$ is linear, it is also true that any linear map can be represented by a matrix.  
To see why this is: consider your example, where you are dealing with the map which takes $\left(\begin{matrix} x_1 \\ x_2 \end{matrix} \right)$ to $\left(\begin{matrix} -x_2 \\ x_1+2x_2 \\ 3x_1 - 4x_2 \end{matrix} \right)$.  In mathematics, we say that this is an example of a map from $\mathbb{R}^3$ (vectors with three components) to $\mathbb{R}^2$ (vectors with two components).  I will now show that every linear map from $\mathbb{R}^3$ to $\mathbb{R}^2$ can be represented by a matrix in this way.  The general case for maps from $\mathbb{R}^m$ to $\mathbb{R}^n$ is treated in exactly the same way.  
First note that we can write any vector $a=\left(\begin{matrix} a_1 \\ a_2 \end{matrix} \right)$ in $\mathbb{R}^2$ as $$a_1\left( \begin{matrix} 1 \\ 0 \end{matrix} \right)+a_2\left( \begin{matrix} 0 \\ 1 \end{matrix} \right)=a_1e_1+a_2e_2$$ and any vector $b= \left( \begin{matrix} b_1 \\ b_2 \\ b_3 \end{matrix} \right)$ in $\mathbb{R}^3$ as $$b_1\left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) + b_2\left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right) + b_3\left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)=b_1e_1+b_2e_2+b_3e_3$$.  Now let $\alpha$ be a linear map from $\mathbb{R}^2$ to $\mathbb{R}^3$.  Then, if $a\in \mathbb{R}^2$, $$\alpha(a)=\alpha(a_1e_1+a_2e_2)=a_1\alpha(e_1)+a_2\alpha(e_2)$$
Now, since for $i=1,2$, $\alpha(i) \in \mathbb{R}^3$, we can write $$\alpha(e_i)=m_{1i}e_1+m_{2i}e_2+m_{3i}e_3$$
where the $m_{ij}$ are the components of the images of the vectors $e_i$ under $\alpha$.  Writing the $m_{ij}$ as a matrix 
$$M=\left( \begin{matrix} m_{11} & m_{12} \\ m_{21} & m_{22} \\ m_{31} & m_{32} \end{matrix} \right)$$
it is quick to check that applying $\alpha$ to the vector $a=\left( \begin{matrix} a_1 \\ a_2 \end{matrix}\right)$ is exactly equivalent to multiplying by the matrix $M$.  
Your question asks the opposite: given a matrix $M$,does multiplication by $M$ always define a linear map.  Your calculations (which are perfectly precise) show that the answer is yes, although it's such a basic result that I think showing that the map can be represented by a matrix is probably enough for that question.  
Moral: for vector spaces over the real numbers, linear maps are matrices are linear maps.  Finding a matrix representation is a great way to show that a map is linear.  
I'm not wure what was intended for this question, but you could show that the map is linear directly (without inventing matrices).  That might be a slightly nicer example, as it doesn't assume properties of matrix multiplication which are non-obvious (though trivial to prove).  
