Let $e_1$, $e_2$, and $e_3$ be a base in $\mathbb{R}^3$. Is there a linear operator $A$ such that $A(e_i + e_j) = e_k$? Let $e_1$, $e_2$, and $e_3$ be a base in $\Bbb R^3$.  Is there a linear operator $A$ such that $A(e_1 + e_2) = e_3$, $A(e_2 + e_3) = e_1$, and $A(e_1 + e_3) = e_2$? If it doesn’t exist, explain why.
$A(e_1+e_2)=A(e_1)+A(e_2)$, same for $A(e_2+e_3)$, $A(e_1 + e_3)$.
Solving the system I get that $A(e_2)=(e_3-e_2+e_1)/2$, $A(e_3)=(e_1-e_3+e_2)/2$, $A(e_1)=(e_3+e_2-e_1)/2$. So now if I take $e_1=f_1$, $e_2=f_2$, $e_3=f_3$, $A(e_2)=(f_3-f_2+f_1)/2$, $A(e_3)=(f_1-f_3+f_2)/2$, $A(e_1)=(f_3+f_2-f_1)/2$. 
So I get that this is possible, but it seems to me that I am not right, correct me, please.
 A: If all you wanted was to know if such a linear map exists, all you needed (as you were told in the comments) was to notice that $\{e_1+e_2,e_2+e_3,e_1+e_3\}$ is a basis of $\mathbb{R}^3$. Then, for $any$ three elements $v_1,v_2,v_3\in\mathbb{R}^3$, there is one and only one linear endomorphism $A$ of $\mathbb{R}^3$ such that $A(e_1+e_2)=v_1$, $A(e_2+e_3)=v_2$, and $A(e_1+e_3)=v_3$.
And if what you want is to know which map is that, you nearly made it. You proved (correctly) that, if such a map existed, it could only be the map such that $A(e_1)=\frac12(-e_1+e_2+e_3)$, that $A(e_2)=\frac12(e_1-e_2+e_3)$, and that $A(e_3)=\frac12(e_1+e_2-e_3)$. Therefore, it is the map defined by$$A(x,y,z)=\frac12(-x+y+z,x-y+z,x+y-z).$$
A: HINT.-By linearity you have
$$A(e_1)+A(e_2)=e_3\\A(e_2)+A(e_3)=e_1\\A(e_3)+A(e_1)=e_2$$ you have then a system
$$\begin{cases}x+y=a\\y+z=b\\z+x=c\end{cases}\Rightarrow\begin{cases} 2x=b+c-a\\2y=a+c-b\\2z=a+b-c\end{cases}$$ The solution is unique  . $A$ being linear it is well defined if the images of a base is known.
