How to show that a mapping is linear Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T\begin{pmatrix}u_1\\u_2\end{pmatrix} = \begin{pmatrix}u^2_2\\2\\u_1-u_2\end{pmatrix}$
Show T is linear.

How can I do this using the rules I know
$i)  \, T(u +v) = T(u) + T(v)$
$ii) \,  T(ru) = rT(u)$ ?
For example, in this case what is $u$ and $v$ for me?
UPDATE

Let $S: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T(\begin{pmatrix}u_1\\u_2\end{pmatrix}) = \begin{pmatrix}u_2^2\\2\\u_1-u_2\end{pmatrix}$. Show $S$ is not linear

So I think I've got a solution but want to double check that this answer will suffice. (By the way I let $T = S$
I know that I can use the property
$T(ru) = rT(u)$ and let $u = 0$ to get $T(0) = \begin{bmatrix}0\\2\\0\end{bmatrix} \ne 0 \implies$ not linear. Isn't that all I really need to prove that $S$ is not linear
To answer your question:

$u,v$ are just elements of a vector space, in your case it is $\mathbb R^2$, that means
$$
u=(u_1,u_2),v=(v_1,v_2) \text{ with } u_1,u_2,v_1,v_2\in \mathbb R
$$
$r$ on the other hand is called a scalar and in your case the corresponding filed (over which your vector space is defined) is again $\mathbb R$, so $r\in \mathbb R$.

The following was an answer to the question before editing:

Now just check for example what @Bye_World already mentioned in the comments to see that 
$$
T(ru) = rT(u)
$$
fails (or check whether the $0$ is mapped to $0$) or just check your first property with 
$$
u=(1,0),v=(0,1)
$$
which gives
$$
T(u+v)=T((1,1))=(1,2,0)\neq(0,2,1)+(1,2,-1)=(1,4,0)=T(u)+T(v)
$$
So $T$ is not a linear mapping.

The following is an answer to the question after editing

Here $T$ is indeed linear, the second property is obvious and for the first one we just check
\begin{align}
T(u+v)=T((u_1+v_1,u_2+v_2))&amp;=(u_2+v_2,0,u_1+v_1-u_2-v_2)\\&amp;=(u_2,0,u_1-u_2)+(v_2,0,v_1-v_2)=T(u)+T(v)
\end{align}
In this case, your $u$ and $v$ will be vector in the form $\begin{pmatrix} a\\b\end{pmatrix}$.

To check the first property, let $u = \begin{pmatrix} a\\b \end{pmatrix},v=\begin{pmatrix} c\\d \end{pmatrix}.$ Then we have

$$T(u+v)=T(\begin{pmatrix} a\\b \end{pmatrix}+\begin{pmatrix} c\\d \end{pmatrix})=T(\begin{pmatrix} a+c \\b+d \end{pmatrix})  \text{......(why?)}$$

You then compute $T(u) + T(v)$. If these two are equal then you are done for the first property. However, I am not saying that $T$ is a linear transformation since I haven't done the computation. But in fact you can tell that $T$ is not a linear transformation. This is a consequence of the following fact:

https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector