# A vector function of vectors

If we are given an arbitrary function $$\vec{\textbf{a}}:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^k$$, what would be an example of this? Say with $$n=2, k=3$$ or something along those lines?

Along the same line, what about a scalar function $$a:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$$? What would this look like?

My guess for $$\vec{\textbf{a}}$$ is something along the lines of $$\vec{\textbf{a}}=\big<\vec{\textbf{x}}-\vec{\textbf{y}}, \vec{\textbf{x}}+\vec{\textbf{y}}\big>$$ if $$n=2, k=4$$, but this seems strange. Having vectors inside of vectors. Does that make this a vector with 4 components? Also, this would be very limiting to what $$k$$ could be, without the dot product to make make entries of the vector be non-multiples of $$n$$.

I'm very unsure about the scalar function, and my only guess is to take a bunch of variables $$x_i$$ and $$y_i$$ and make a normal function with $$2n$$ scalar variables, but that seems wrong, because then it would just be $$a:\mathbb{R}^{2n} \to \mathbb{R}$$, and there would be no reason to have the cartesian product in there.

So, what would these functions look like?

## 1 Answer

Just as $$\mathbb R\times \mathbb R=\mathbb R^2$$ we have that $$\mathbb R^n\times\mathbb R^n=\mathbb R^{2n}$$. It seems that you are thinking of $$\mathbb R^2\times\mathbb R^2$$ as a set consisting of elements of the form $$((a,b),(c,d))$$, which is technically true, but this is equivalent to the set consisting of elements of the form $$(a,b,c,d)$$ by identifying the elements of each set in a one-to-one fashion with elements of the other in the natural way. You can think of this as the associativity of the Cartesian product: just like it doesn't matter how you group elements in a summation, it doesn't matter how you group coordinates in a Cartesion product. In effect "vectors inside vectors" are the same as bigger vectors.

Thus any function $$f:\mathbb R^n\times \mathbb R^n\to \mathbb R$$, defined by $$\mathbb R^n\times \mathbb R^n\ni(\mathbf u,\mathbf v)\mapsto f(\mathbf u,\mathbf v)$$, can be identified with a $$g:\mathbb R^{2n}\to \mathbb R$$ given by $$\mathbb R^{2n}\ni(u_1,\dots,u_{2n})\mapsto f((u_1\dots,u_n),(u_{n+1},\dots u_{2n}))$$. Naturally this identification goes the other way, so these two classes of functions are the same, with the only difference being the number of brackets needed. Hence for your purposes it is simpler to think of $$\mathbb R^n\times \mathbb R^n$$ as $$\mathbb R^{2n}$$, and define the infinite variety of functions in the usual pointwise manner.