Answer
We verify properties (1) and (2), and show that any map satisfying these properties is linear.
Work Step by Step
Part 1. We verify properties (1) and (2) for linear functions
Let $G$ be a linear function in the form:
$G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
Evaluate $G\left( {{u_1} + {u_2},{v_1} + {v_2}} \right)$:
$G\left( {{u_1} + {u_2},{v_1} + {v_2}} \right) = \left( {A\left( {{u_1} + {u_2}} \right) + C\left( {{v_1} + {v_2}} \right),B\left( {{u_1} + {u_2}} \right) + D\left( {{v_1} + {v_2}} \right)} \right)$
$ = \left( {A{u_1} + C{v_1} + A{u_2} + C{v_2},B{u_1} + D{v_1} + B{u_2} + D{v_2}} \right)$
$ = \left( {A{u_1} + C{v_1},B{u_1} + D{v_1}} \right) + \left( {A{u_2} + C{v_2},B{u_2} + D{v_2}} \right)$
Since $G\left( {{u_1},{v_1}} \right) = \left( {A{u_1} + C{v_1},B{u_1} + D{v_1}} \right)$
and $G\left( {{u_2},{v_2}} \right) = \left( {A{u_2} + C{v_2},B{u_2} + D{v_2}} \right)$, so
$G\left( {{u_1} + {u_2},{v_1} + {v_2}} \right) = G\left( {{u_1},{v_1}} \right) + G\left( {{u_2},{v_2}} \right)$
This verifies property (1).
Evaluate $G\left( {cu,cv} \right)$ for any constant $c$:
$G\left( {cu,cv} \right) = \left( {A\left( {cu} \right) + C\left( {cv} \right),B\left( {cu} \right) + D\left( {cv} \right)} \right)$
$ = c\left( {Au + Cv,Bu + Dv} \right)$
So, $G\left( {cu,cv} \right) = cG\left( {u,v} \right)$.
This verifies property (2).
Part 2. We show that any map satisfying properties (1) and (2) is linear.
Let $G$ be a map such that $G\left( {u,v} \right) = \left( {\phi \left( {u,v} \right),\psi \left( {u,v} \right)} \right)$, where $\phi $ and $\psi $ are functions of $u$ and $v$.
Suppose that $G$ is a map that satisfies properties (2) such that
$G\left( {cu,cv} \right) = cG\left( {u,v} \right)$
By definition, the left-hand side and the right-hand side become:
$\left( {\phi \left( {cu,cv} \right),\psi \left( {cu,cv} \right)} \right) = c\left( {\phi \left( {u,v} \right),\psi \left( {u,v} \right)} \right)$
$\left( {\phi \left( {cu,cv} \right),\psi \left( {cu,cv} \right)} \right) = \left( {c\phi \left( {u,v} \right),c\psi \left( {u,v} \right)} \right)$
Thus,
$\phi \left( {cu,cv} \right) = c\phi \left( {u,v} \right)$, ${\ \ \ \ }$ $\psi \left( {cu,cv} \right) = c\psi \left( {u,v} \right)$
This implies that $\phi $ and $\psi $ are linear functions of $u$ and $v$. So, we can write:
$\phi \left( {u,v} \right) = Au + Bv$, ${\ \ \ \ }$ $\psi \left( {u,v} \right) = Cu + Dv$,
where $A$, $B$, $C$, and $D$ are constants.
Thus, $G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
In part (1) above we have shown that $G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$ satisfies properties (1). By definition of linearity, $G$ is a linear map.
Hence, we conclude that any map satisfying properties (1) and (2) is linear.
