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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](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Most of what you've written is already in proper form. All you have to do is wrap it in dollar signs.2012-09-11
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    You need to show that (i) $T(\vec{x}+\vec{y}) = T(\vec{x})+T(\vec{y})$, and (ii) $T(\lambda \vec{x}) = \lambda T(\vec{x})$.2012-09-11
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    @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.2012-09-11

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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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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.2012-09-11
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    @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 :)2012-09-11
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    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)?2012-09-11
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    @BMEdwards37 The $x$ is just a generic vector.. you are free to pick another generic vector named $y$ and having coordinates that you call $y_1, y_2, y_3$.2012-09-11
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    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.2012-09-11
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    @BMEdwards37 You *don't* only have $x$. To verify the axioms you *have* to posit two generic vectors. Did you try to compute what $T((x_1,x_2,x_3)+(y_1,y_2,y_3))$ is? I was waiting for that...2012-09-11
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    I'm sorry, I had moved on but now i guess I am back - I was just getting frustrated and i'm wasting a *lot* of time. But if i just pick two random vectors, say (1, 2, 3) and (3, 2, 1) and do T(x+y) and T(x)+T(y) along with trying one of them with a scalar, i will have proved it?2012-09-11
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    @BMEdwards37 No. You will have proved it for just one pair of vectors. If this is all you do, then you might have "gotten lucky" and randomly picked two vectors that *just happen* to work. To prove it in general, you need to use two generic vectors, as I wrote. You seem pretty reluctant to do that... but it's not hard! Give it a shot. I also added in a fragment that I accidentally left out.. it should be OK now.2012-09-11
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    I see what you're saying now, it's always the simple things. Thank you for the help.2012-09-11