Aditivity of a linear mapping question. We have a linear mapping $A: (x,y) \to (x, e^y)$.
When proving additivity I have a doubt. We have to prove that $A(\vec{x}+\vec{y}) = A\vec{x} + A\vec{y}$. Let $\vec{x} = (x,y)$ and $\vec{y} = (x',y')$.
We go like this: $A(\vec{x}+\vec{y})$ = $A((x,y) + (x',y'))$ = $A(x + x', y + y')$ = $(x + x', e^y + e^{y'})$ = $(x, e^y) + (x', e^{y'})$ = $A\vec{x} + A\vec{y}$.
My question: When we evaluate this $A(x + x', y + y')$ do we take component by component, $x \to x$ and $x' \to x'$, so $x + x' \to x + x'$ or we take the  whole first component in this case $x + x'$ and we see how the mapping acts on the first component, which in this case is the same. But when it comes to the second component it can be either $e^y + e^{y'}$, if we consider component by component or $e^{y+y'}$ if we consider the whole component.
How we evaluate this, I guess the answer would apply also to proving homogeneity.
Thanks a lot.
 A: x part is linear however y part is not. Consider this:
$$A(y)=e^y$$
$$A(y_1)+A(y_2)=e^{y_1}+e^{y_2}$$
$$A(y_1+y_2)=e^{y_1+y_2}$$
so clearly
$$A(y_1+y_2)\neq A(y_1)+A(y_2)$$
as a sanity check, I usually just compare what happens when I insert $2y$, it is best to do it the long way to be sure but if it doesn't work for $2y$, you can immediately rule out linearity.
$$e^{2y}\neq 2e^y$$
EDIT: I would also like to talk about something extra, because so many people get confused. What about this:
$$A(\ y(x)\ )=x^2y(x)$$
This is still linear.
$$A(\ y_1(x)\ )+A(\ y_2(x)\ )=x^2y_1(x)x^2y_2(x)=x^2(y_1(x)+y_2(x))$$
$$A(\ y_1(x)\ y_2(x)\ )=x^2(y_1(x)+y_2(x))$$
$$A(\ y_1(x)\ )+A(\ y_2(x)\ )=A(\ y_1(x)\ y_2(x)\ )$$
A: $A(x,y)=(x,e^y)$ means you map the $x$ component of the vector to itself, $y$ component to $e^y$.
If your vector now is $(x+x',y+y')$, $A$ maps it to $(x+x', e^{y+y'})$ because $y+y'$ is the $y$ component of this whole vector.
It is then not linear.
A: If $A$ is a linear map, then $A(\vec{0})=\vec{0}$ (by homogeneity). In your case,
$$
A(\vec{0})=A(0,0)=(0,e^0)=(0,1)\ne\vec{0}
$$
so you rule out linearity.
However, this is just a good shortcut in some cases; the map $B(x,y)=(x,y^2)$ satisfies $B(\vec{0})=\vec{0}$, but is not linear. Try with homogeneity with $\alpha=-1$ and $\vec{x}=(1,-1)$, showing that $B(\alpha\vec{x})\ne\alpha B(\vec{x})$.
In general, in order to show non linearity, you have to show a counterexample. The ones I used for $A$ and $B$ are sufficient. I suggest you to first attack with $\vec{0}$, then with homogeneity and only as a last weapon with additivity.
