the orbit of an element and a set of inner products 
Let $i : \mathbb{R}^n \times \mathbb{R}^n\to \mathbb{R} : (x,y)\mapsto x^T y$ be the standard inner product on $\mathbb{R}^n.$ Let $GL_n(\mathbb{R})$ act on $\mathbb{R}^{\mathbb{R}^n \times \mathbb{R}^n },$ where $A^B$ is the set of functions $f : B\to A,$ with the diagonal action $\cdot_1$ on $\mathbb{R}^n\times \mathbb{R}^n$ and the trivial action on $\mathbb{R},$ so that for $A \in GL_n(\mathbb{R}), f \in \mathbb{R}^{\mathbb{R}^n \times \mathbb{R}^n }, (A\cdot f)(x_1,x_2) = f(A^{-1}\cdot_1 (x_1,x_2)).$ Show that the orbit of $i$ in $\mathbb{R}^{\mathbb{R}^n \times \mathbb{R}^n }$ is the set of inner products on $\mathbb{R}^n.$

I think I can show that an element of the orbit is an inner product using a fairly tedious method (checking all the inner product axioms and using the definition of the action and the definition an orbit). However, I'm not sure how to show that the set of inner products is contained in the orbit of $i$. I know that for $A \in GL_n(\mathbb{R}), f \in \mathbb{R}^{\mathbb{R}^n \times \mathbb{R}^n }, (A\cdot i)(x_1, x_2) = x_1^T(A^{-1})^TA^{-1}x_2.$ But I'm not sure how to proceed from here.
 A: You want to show that any inner product is of the form $(x,y) \mapsto x^T \, B \, y$ for some matrix $B$. You can do this by choosing a basis and then using the properties of an inner product.
If you have a basis $e_1,\dots,e_n$, then write the vectors in this basis as
$$x = (x_1,\dots,x_n) = \sum_{i=1}^n x_i e_i \quad \text{and} \quad y=(y_1,\dots,y_n) = \sum_{i=1}^n y_i e_i$$

*

*One of the main properties of an inner product is linearity. Let $\left<x,y\right>$ denote the inner product. Then
$$ \left<x,y\right> = \left< \sum_i x_i e_i, \; \sum_j y_j e_j \right> = \sum_{i,j} x_iy_j \left<e_i,e_j\right> $$
Let $B$ be the matrix with entries $b_{ij} = \left<e_i,e_j\right>$. Then it is easy to check that $\left<x,y\right> = x^T \, B \, y$ (just do the matrix multiplication and you will get the expression above!)


*Another property of an inner product is symmetry: $\left<x,y\right> = \left<y,x\right>$. You can convince yourself that this means the matrix $B$ must be a symmetric matrix.


*An inner product is also positive definite. It is known that any symmetric positive definite matrix can be decomposed as $A^T A$ for some matrix $A$. I'm sure you can find many posts on this site about why this is true (for example this question)
