Help determining whether a transformation is linear or not I have the following transformation:

$T : \Bbb R^3 → \Bbb R^2 , T (x_1 , x_2 , x_3 ) = (x_1 − x_2 , 2x_2 )$

I need to determine whether it's a linear transformation or not.
I understand that for a transformation to be linear, I should have:
\begin{equation*}
T(av + bu) = aT(v) + bT(u)
\end{equation*}
Now for the question above let's say I do the following:
$u = ax_1 + ax_2 + ax_3$
$v = bx_1 + bx_2 + bx_3$
So now, I have to show that $T(au+bv) = aT(u) + bT(v)$. I'm quite stuck, and I'm not sure how to proceed. Could anyone guide me on where I'm wrong and how to proceed?
Thanks.
 A: $v=(x_1,x_2,x_3)$ & $w=(y_1,y_2,y_3)$
We have then
$$T(av+bw)=(ax_1+by_1-(ax_2+by_2),2(ax_2+by_2))=$$
$$=(ax_1+by_1-ax_2-by_2,2ax_2+2by_2)$$
$$=(ax_1-ax_2+by_1-by_2,2ax_2+2by_2)$$
$$=(ax_1-ax_2,2ax_2)+(by_1-by_2,2by_2)$$
$$=(a(x_1-x_2),2ax_2)+(b(y_1-y_2),2by_2)$$
$$=a(x_1-x_2,2x_2)+b(y_1-y_2),2y_2)$$
$$=aT(v)+bT(w)$$
A: My first inclination for your approach would be to do exactly as Zelos Malum has done in their answer. 
Alternatively, instead of trying to tackle the "linear-combination-preservation" property all at once for a transformation to determine whether it's linear, it might be helpful just to break that property down into 


*

*Closure under vector addition Here you need to show that $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$.

*Closure under scalar multiplication (where your scalar is defined over the appropriate field ($\mathbb{Q}, \mathbb{R}, \mathbb{C}$, etc.). This means you need to show that if $k$ is a scalar, then $T(k\vec{v}) = kT(\vec{v})$.


So I'll demonstrate by letting $\vec{u}, \vec{v} \in \mathbb{R}^3$ such that $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$. Then we have
$$ T(\vec{u} + \vec{v}) \\
= T((u_1 + v_1), (u_2 + v_2), (u_3 + v_3)) \\
= ((u_1 + v_1) - (u_2+v_2), 2(u_2+v_2)) \\
= ((u_1-u_2)+(v_1-v_2), 2u_2+2v_2) \\
= (u_1-u_2, 2u_2) + (v_1-v_2, 2v_2) \\
= T(\vec{u}) + T(\vec{v}).
$$
Thus, we can conclude that $T$ is closed under vector addition.
Next, let $k$ be a scalar from the same field as the components of $\vec{v}$; that is, let $k \in \mathbb{R}$. Then we have
$$
T(k\vec{v}) \\
= T((kv_1, kv_2, kv_3)) \\
= (kv_1-kv_2, 2kv_2) \\
= (k(v_1-v_2), k(2v_2))\\
= k(v_1-v_2, 2v_2) \\
= kT(\vec{v}).
$$
This means that $T$ is closed under scalar multiplication. This logically equivalent to saying that $T$ preserves linear combinations of vectors and so $T$ is linear.
A: You would do better to say 
$\mathbf{v} = (v_1,v_2,v_3)$
$\mathbf{u} = (u_1,u_2,u_3)$
so you can say $a\mathbf{v}+b\mathbf{u} = (av_1+bu_1, av_2+bu_2, av_3+bu_3)$.
Now state what $T(a\mathbf{v}+b\mathbf{u})$ is and what $aT(\mathbf{v})+bT(\mathbf{u})$ is and whether they are equal.
