Show that a reflection of each vector $\vec{x}=(x_1, x_2, x_3)$ through $x_3=0$ onto $T(\vec{x})=(x_1, x_2, -x_3)$ is linear. 
Show that a reflection of each vector $\vec{x}=(x_1, x_2, x_3)$ through $x_3=0$ onto $T(\vec{x})=(x_1, x_2, -x_3)$ is linear.

I think it somehow involves the Transformation Matrix: $A=\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&-1\end{bmatrix} \times \vec{x}$, resulting in $(x_1, x_2, -x_3)$ but i'm not sure. I'm wondering if it has something to do with the superposition principle but i'm also not sure.
 A: You need to show that (see this)
$$
T\left(a_1 {\bf v}_1 +  a_2 {\bf v}_2\right) = a_1 T\left({\bf v}_1\right) + a_2 T\left({\bf v}_2\right)
$$
for $T\left(v_x,v_y,v_z\right) = \left(v_x,v_y,-v_z\right)$, with general scalars $a_1, a_2$ and general vectors ${\bf v}_1 = \left(v_{1x}, v_{1y}, v_{1z}\right)$ and ${\bf v}_2 = \left(v_{2x}, v_{2y}, v_{2z}\right)$.
Start with the left hand side:
$$
\begin{eqnarray}
T\left(a_1 {\bf v}_1 +  a_2 {\bf v}_2\right) &=& T\left(a_1\left(v_{1x}, v_{1y}, v_{1z}\right) + a_2\left(v_{2x}, v_{2y}, v_{2z}\right)\right) \\
&=& T\left(\left(a_1 v_{1x}, a_1 v_{1y}, a_1 v_{1z}\right) + \left(a_2 v_{2x}, a_2 v_{2y}, a_2 v_{2z}\right)\right) \\
&=& T\left(a_1 v_{1x} + a_2 v_{2x}, a_1 v_{1y} + a_2 v_{2y}, a_1 v_{1z} + a_2 v_{2z}\right) \\
&=& \left(a_1 v_{1x} + a_2 v_{2x}, a_1 v_{1y} + a_2 v_{2y}, - a_1 v_{1z} - a_2 v_{2z}\right) \\
&=& \left(a_1 v_{1x}, a_1 v_{1y}, -a_1 v_{1z}\right) + \left(a_2 v_{2x}, a_2 v_{2y}, -a_2 v_{2z}\right) \\
&=& a_1 \left(v_{1x}, v_{1y}, -v_{1z}\right) + a_2 \left(v_{2x}, v_{2y}, -v_{2z}\right) \\
&=& a_1 T\left({\bf v}_1\right) + a_2 T\left({\bf v}_2\right).
\end{eqnarray}
$$
A: Your matrix $A$ is correct, and as you are probably well aware, mulitplying a tuple with a matrix is a linear transformation.
If you like to (or need to, depending on your confidence with the above fact), you can go and verify all the axioms:
Show that $T((x_1,x_2,x_3)+(y_1,y_2,y_3))=T(x_1,x_2,x_3)+T(y_1,y_2,y_3)$
and $T(\alpha(x_1,x_2,x_3))=\alpha T((x_1,x_2,x_3))$
