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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.

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LaTeX is a nice touch, but don't let it stop you from doing anything. If you're looking for a crash course we have this. Most of what you've written is already in proper form. All you have to do is wrap it in dollar signs. –  axblount Sep 11 '12 at 17:12
    
You need to show that (i) $T(\vec{x}+\vec{y}) = T(\vec{x})+T(\vec{y})$, and (ii) $T(\lambda \vec{x}) = \lambda T(\vec{x})$. –  copper.hat Sep 11 '12 at 17:15
    
@axblount ah i see, i was trying the $ in the beginning but missed that i had to have one at the end so I just got rid of it. –  BMEdwards37 Sep 11 '12 at 17:35

2 Answers 2

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 $T(\alpha(x_1,x_2,x_3))=\alpha T((x_1,x_2,x_3))$

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I do know that it's linear, but I have to justify it and that's where my issue comes in. I can write it in $x_1v_1+x_2v_2+x_3v_3$ vector notation, but I gap on how i prove the two properties of linearity - (T(x+y)=T(x)+T(y)) and superposition. –  BMEdwards37 Sep 11 '12 at 17:34
    
@BMEdwards37 You don't need to resort to putting $v$'s in if you're going to identify everything with coordinates and the transformation matrix. You can add and scale vectors, right? Just confirm the two conditions I wrote :) –  rschwieb Sep 11 '12 at 17:39
    
Also, i only have vector x with $x=(x_1, x_2, x_3)$ and no Y, so where does that come up in T(x+y)? –  BMEdwards37 Sep 11 '12 at 17:40
    
@BMEdwards37 The $x$ is just a generic vector.. you are free to pick another generic vector named $y$ and having coordinates that you call $y_1, y_2, y_3$. –  rschwieb Sep 11 '12 at 17:40
    
Thanks for the help, i understand exactly what you're trying to say but have absolutely no idea exactly how to do it with my generic case where i only have $x=(x_1, x_2, x_3)$ so i guess i'll just skip it and move on before i become a bother, i've already wasted over two hours on this problem. –  BMEdwards37 Sep 11 '12 at 17:47

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: $$ \begin{eqnarray} T\left(a_1 {\bf v}_1 + a_2 {\bf v}_2\right) &=& T\left(a_1\left(v_{1x}, v_{1y}, v_{1z}\right) + a_2\left(v_{2x}, v_{2y}, v_{2z}\right)\right) \\ &=& T\left(\left(a_1 v_{1x}, a_1 v_{1y}, a_1 v_{1z}\right) + \left(a_2 v_{2x}, a_2 v_{2y}, a_2 v_{2z}\right)\right) \\ &=& T\left(a_1 v_{1x} + a_2 v_{2x}, a_1 v_{1y} + a_2 v_{2y}, a_1 v_{1z} + a_2 v_{2z}\right) \\ &=& \left(a_1 v_{1x} + a_2 v_{2x}, a_1 v_{1y} + a_2 v_{2y}, - a_1 v_{1z} - a_2 v_{2z}\right) \\ &=& \left(a_1 v_{1x}, a_1 v_{1y}, -a_1 v_{1z}\right) + \left(a_2 v_{2x}, a_2 v_{2y}, -a_2 v_{2z}\right) \\ &=& a_1 \left(v_{1x}, v_{1y}, -v_{1z}\right) + a_2 \left(v_{2x}, v_{2y}, -v_{2z}\right) \\ &=& a_1 T\left({\bf v}_1\right) + a_2 T\left({\bf v}_2\right). \end{eqnarray} $$

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