Determining whether maps $T:\mathbb{R}^3 \rightarrow\mathbb{R}^2$are linear transformation The questions are to determine which one of these 3 maps $T:\mathbb{R}^3 \rightarrow\mathbb{R}^2$are linear transformation. I know that it's the case of $T(\alpha v_1+\beta v_2) = \alpha T(v_1)+\beta T(v_2)$ for all vectors $v_1$ and $v_2$, and all numbers $\alpha$ and $\beta$.
Question 1:
$$
T=
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        x+y+z\\
        xyz \\
        \end{matrix}
        \right)
$$
My last line of calculation to Q1:
$$
        \alpha
        \left(
        \begin{matrix}
        \alpha x+2x\beta u+\alpha y^2+2y\beta u\\
        2x-y \\
        \end{matrix}
        \right) - 
        \beta
        \left(
        \begin{matrix}
        \beta u+\beta v\\
        2u+v \\
        \end{matrix}
        \right)
$$
Question 2:
$$T=
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        x^2+y^2\\
        2x-y \\
        \end{matrix}
        \right)
$$
My last line of calculation for Q2:
$$
\alpha
        \left(
        \begin{matrix}
        \alpha x^2 + \alpha y^2 +2x\beta u+2y\beta v\\
        2x-y \\
        \end{matrix}
        \right) + 
        \beta
        \left(
        \begin{matrix}
        u^2+v^2\\
        2u-v \\
        \end{matrix}
        \right)
$$
Question 3:
$$T=
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        (x+1)^2-(x^2+1)\\
        x+y \\
        \end{matrix}
        \right)
$$
My last line of calculation for $$
\alpha
        \left(
        \begin{matrix}
        2x\\
        x+y \\
        \end{matrix}
        \right) + 
        \beta
        \left(
        \begin{matrix}
        2u\\
        u+v \\
        \end{matrix}
        \right)
$$
I found that all are not linear but I'm a little worried that I may be wrong in one of these questions. Can you help me clarify my answers? Sorry for the messy format. If you have an idea on how to make it look better, feel free to edit the appearance.
Additional question:
$$T=
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        3\\
        3 \\
        \end{matrix}
        \right)
$$
I would say this is linear but I may be wrong for this one as well. Can anyone validate my answer?
 A: Your first transformation is not linear since 
$$(6,8)=T(2,2,2)=T((1,1,1)+(1,1,1))\ne T(1,1,1)+T(1,1,1)=(3,1)+(3,1)=(6,2).$$ Note that the presence of $xyz$ makes it not linear.
The same happens to the second transformation. In this case the presence of $x^2+y^2$ makes it nonlinear. Note that
$$(8,2)=T(2,2,2)=T((1,1,1)+(1,1,1))\ne T(1,1,1)+T(1,1,1)=(2,1)+(2,1)=(4,2).$$
However note that the third one can be written as
$$T
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        (x+1)^2-(x^2+1)\\
        x+y \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        2x\\
        x+y \\
        \end{matrix}
        \right)$$
and it is linear. Can you show it?
Note that a linear transformation must have the form
$$T
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        ax+by+cz\\
        dx+ey+fz \\
        \end{matrix}
        \right)$$ where $a,b,c,d,e,f$ are real numbers. If there is an expression (after simplification, of course) as $x^2$ or $xy$ or $5$ then it is not linear.
A: You're correct in thinking that neither of these are linear transformations. However, the best way to convince a skeptic that this is the case is to give specific examples that contradict linearity.
For both questions consider $2\cdot T(1,1,1)$ and check if it equals $T(2,2,2)$. If it doesn't, then you're done. 
A: Re your additional question,
$$
T
        \left(
        \begin{matrix}
        x \\
        y \\
        z \\
        \end{matrix}
        \right)=
\left(
        \begin{matrix}
        3\\
        3 \\
        \end{matrix}
        \right)
$$
is not linear, because a linear transformation preserves the origin, since for any linear transformation $A$:
$$ 
A(\mathbf{0})=A(\mathbf{x}-\mathbf{x})=A(\mathbf{x})-A(\mathbf{x})=0
$$
Being linear is a very restrictive property since it preserves the structure of a vector space and in this case a constant function would just contract the vector space to a single point. (However note, that $B(\mathbf{x})=0$ is linear since $0 = B\left(\alpha \mathbf{x} + \beta \mathbf{y}\right)=\alpha B(\mathbf{x}) + \beta B(\mathbf{y}) = 0 + 0$)
