When does a matrix define a metric on $\mathbb{R}^n$ Given a positive definite matrix $A\in\mathbb{R}^{n\times n}$, I need to construct a metric on $\mathbb{R}^n$ using $A$ that is equivalent to the Euclidean metric $\|x-y\|$.

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*First I thought maybe $\langle A(x-y),x-y \rangle $, would this work, please let me know?


*Next I thought $\sqrt{\langle A(x-y),x-y \rangle}$, since
$$C_1\|v\|^2 \leq  \langle Av,v \rangle\leq C_2 \|v\|^2 \implies \sqrt{C_1}\|v\| \leq  \langle Av,v \rangle\leq \sqrt{C_2} \|v\|. $$
 A: Your second idea is the good one. The first just lacked the square root. Here is a proof.
Define $\|x\|_A^2=\langle Ax,x \rangle$.
$A$ is positive definite, hence $A=ODO^T$ with $O$ an orthogonal matrix and $D$ a diagonal matrix, having only positive elements on the diagonal.
$$\|x\|_A^2=\langle Ax,x \rangle=\langle ODO^Tx,x \rangle=\langle OD^{1/2}x,OD^{1/2}x \rangle$$
where $D^{1/2}$ is the matrix of the square root of $D$ element-wise ( which is well defined since all values are non-negative).
Since $\| x\|^2=\langle x,x\rangle$, finally :
$$\|x\|_A^2=\| OD^{1/2}x \|^2=\|D^{1/2}x \|^2$$
Last equality holds since $O$ is orhogonal.
Finally, if we note $\lambda_1,\dots,\lambda_n$ the diagonal elements of $D$, we have
$$\|x\|_A^2=\sum_{k=1}^n |\sqrt{\lambda_k}x_k|^2$$
According to this expression, $\| \cdot \|_A$ is clearly a norm. Every norm on finite dimension space are equivalent to each other. However here are the explicit bounds  for that norm
$$\min_k \sqrt \lambda_k \| x\| \le \|x\|_A \le \max_k \sqrt \lambda_k \| x\|$$
where $\lambda_k$ are recalled to be the eigenvalues of $A$
