Proving that two metric spaces are isometric

Question

I think (hope) I'm on the right track with this problem, but there are details that I can't seem to work out. I've also struggled to find examples of this sort of problem to assist me.

Let
$\rho (\mathbf{x}, \mathbf{y}) = \sqrt{(\mathbf{x} - \mathbf{y})^T \mathbf{A} (\mathbf{x} - \mathbf{y})}$,
where $\mathbf{x}=(x_1,\ldots,x_n)$, $\mathbf{y}=(y_1,\ldots,y_n)$, and $\mathbf{A}$ is an $n \times n$ symmetric positive-definite real matrix. Taking $\varepsilon$ to be the ordinary Euclidean metric, show that $(\mathbb{R}^n,\varepsilon)$ and $(\mathbb{R}^n,\rho)$ are isometric.

My thought process so far: I need to show that $\exists f:\mathbb{R}^n \rightarrow \mathbb{R}^n$, which is surjective, such that
$\rho(f(\mathbf{x}),f(\mathbf{y})) = \varepsilon(\mathbf{x},\mathbf{y})$.
What I'm thinking is to use Sylvester's Law of Inertia, so that
$(f(\mathbf{x})-f(\mathbf{y}))^T \mathbf{A} (f(\mathbf{x})-f(\mathbf{y})) = \sum_{i=1}^n w_i^2$.
This would reduce the equation to $\sum_{i=1}^n w_i^2 = \sum_{i=1}^n (x_i - y_i)^2$, but I'm too tentative to assert that $w_i = x_i-y_i$ for some $f$.

How should I proceed? Or am I completely off-track here? Please be brutal.

Answer 1

Positive definite real symmetric matrix must be of the form $\mathbf{A}=\mathbf{P}\mathbf{P}^T$ for some regular matrix $\mathbf P$. (Since it's diagonal form is the identity matrix $\mathbf{I}$. The matrix $\mathbf{P}$ can be even chosen to be orthogonal.)

Now define $f(\mathbf{x})=(\mathbf{P}^T)^{-1}\mathbf{x}$

$\rho (f(\mathbf{x}), f(\mathbf{y})) = \sqrt{(\mathbf{x} - \mathbf{y})^T \mathbf{P}^{-1} \mathbf{P} \mathbf{P}^T (\mathbf{P}^T)^{-1} (\mathbf{x} - \mathbf{y})} = \sqrt{(\mathbf{x} - \mathbf{y})^T (\mathbf{x} - \mathbf{y})} = \varepsilon(\mathbf x, \mathbf y)$.

This is probably very similar to what you wanted to do, but I did not quite exactly follow your last conclusion.