Is the set $\theta = \{((x,y),(3y,2x,x+y)):x,y \in R \}$ a function? If so, what is the domain and range?
I know a set is a function if it is a relation and one input has exactly one output for all inputs but I’m confused with the notation. I know that $xRy$ since $(x,y) \in R \times R$ but $(3y, 2x, x+y)$ is tripping me up since that is a $3$-dimensional coordinate. Any help?
On
Your function takes $(x,y)$ into $(3y,2x,x+y)$ thus the domain is $\mathbb {R^2}$ and the codomain is $\mathbb {R^3}$
It is well defined because if $$(x_1,y_1)=(x_2,y_2)$$ then $x_1=x_2$ and $y_1=y_2$ therefore, $$(3y_1,2x_1,x_1+y_1)= (3y_2,2x_2,x_2+y_2)$$
Your function is indeed a linear transformation which could be represented by a $3\times 2 $ matrix.
Well, that function takes pairs of real numbers $(x,y)$ into $(3y,2x,x+y)$. Thus, the domain of that function, $\theta$, is $\mathbb R^2$ and the codomain is $\mathbb R^3$. The range of $\theta$ is given by $$\{ (3y,2x,x+y):\, x,y\in \Bbb R \} \subset \Bbb R^3$$ As you can see, the function can be written using a more recognizable way instead of a set : $$\begin{align} \theta : \Bbb R^2 &\to \Bbb R^3 \\ (x,y) &\mapsto (3y,2x,x+y) \end{align}$$ or simply $\theta (x,y) = (3y,2x,x+y)$.
Remark : To prove that $\theta$ is a true function you have to use the definition, that is, if $(x_1,x_2)=(y_1,y_2)$ then $\theta (x_1,x_2)= \theta(y_1,y_2)$.