Why is this False: The map $L:\mathbb{R}^2 \to \mathbb{R}$ given by $L(x_1,x_2)=|x_1-x_2|$ is linear Every time I try to prove it not linear though showing it isn't closed under multiplication or addition I keep showing its linear(True). When I know it should be false what am I missing.
Proof-
The Linear map from $\mathbb R^2 \to\mathbb R$ is not linear as it isn't closed under addition. Assume,$x=(1,1)=0$ and $y=(1,0)$ Meaning, $L(x)=|1-1|=0$ and $L(y)=|1-0|=1$ So, $L(x)+L(y)=1$. Then $L(x + y) = L(1,1) + L(1,0) = |1 - 1| + |1 - 0| = 0 + 1 = 1$ This shows that the condition 1 product is satisfied and $L(x_1,x_2) = |x_1 - x_2| $ is a linear map.

In order to be linear, $L(x+y)=L(x)+L(y) $ must hold for all $x$ and $y$.  To prove $L$ is non-linear, it therefore suffices to find some $x$ and $y$ for which  $L(x+y)=L(x)+L(y) $ does not hold.  The map $T(x)=x_1-x_2$ actually is linear, but the map $L(x)=\lvert x_1-x_2\rvert$ with the absolute value is not, so choosing $x$ and $y$ with $x_1>x_2$ but $y_1<y_2$, such as $x=(1,-1)$ and $y=(-1,1)$, works to provide a counterexample to the linearity of $L$.
