Linear algebra: Proof of number of solutions in a linear system. What is going on? In my linear algebra book I see this:

I am confused as to why the nonzero column matrix of $x_b = x_1 - x_2$ is a solution but not the zero one? I am also confused as to why $A(x_1 - x_2) = O$
For context, I just read the passage below and I see how the column matrices don't have to be zero matrices to be a solution to the equation $Ax = 0$:

For example, the matrix:
$$
  \left[ {\begin{array}{c}
   1\\
   4\\
   7\\
  \end{array} } \right]
$$
is a non zero matrix but is a solution to the above example. But in the proof at the beginning, why does $A(x_1 - x_2) = 0$ and why is the nonzero column matrix a solution to the homogeneous system of linear equations Ax = 0? I might need an example to elucidate this concept. How do you know the column matrix is non zero?
Lastly, what is the proof doing?
BTW, I understand homogeneous to mean:

 A: Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$  
Since $x_1\ne x_2$  we get $x_h=x_1-x_2 \ne 0$  
A: Solutions of nonhomogeneous equations have the form
$$
\begin{bmatrix} {\rm any \; solution \;of} \\ {Ax=b} \end{bmatrix} = \begin{bmatrix} {\rm another \; solution \;of} \\ {Ax=b} \end{bmatrix} + \begin{bmatrix} {\rm some \; solution \;of} \\ {Ax=0} \end{bmatrix}.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 0  \end{bmatrix}
$$ 
It has many solutions. For example,
$$
\begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \begin{bmatrix} 5 \\ 0  \end{bmatrix}, \quad \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 10 \end{bmatrix}, \quad \begin{bmatrix} x_3 \\ y_3 \end{bmatrix} = \begin{bmatrix} 5 \\ -200 \end{bmatrix}.
$$ 
But what is the structure of these solutions? One can see that they all have the form
$$
\begin{bmatrix} x_n \\ y_n \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ d \end{bmatrix}.
$$
It turns out (and your book gives you a proof!) that the second term $\begin{bmatrix} 0 \\ d \end{bmatrix}$ is the common form of all the solutions of the corresponding homogeneous equation 
$$
\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0  \end{bmatrix}.
$$ 
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($\begin{bmatrix} 5 \\ 0 \end{bmatrix}$) and adding some solution of the homogeneous one ($\begin{bmatrix} 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 10 \end{bmatrix}$,$\begin{bmatrix} 0 \\ -200 \end{bmatrix}$).
If this answer is not enough, please ask more precisely.
