How to visualise an anisometric (aka weighted Euclidan) space? Let's say $S$ is a $2$-dimensional vector space over the field of the real numbers with a weighted inner product
$$\langle x, y \rangle = x^\top Q y ,$$
where $Q$ is a positive-definite matrix. I want to "visualise" this space, so I can make sense of it.
When $Q$ is diagonal, $S$ is simply a version of $\mathbb{R}^2$ stretched along the Cartesian axes. This is easy to understand, so from now on assume $Q$ is not diagonal.
When $Q$ is not diagonal, it gets complicated, because two vectors that are perpendicular in $\mathbb{R}^2$, such as the standard basis vectors $e_1$ and $e_2$, are not perpendicular in the non-diagonal metric $Q$. In short, I have no idea how to make a mental picture of these vectors, or any other vector, in $S$.
One idea I have is to represent a vector $x$ in $\mathbb{R}^2$ along with a transformation $Tx$ which has the length and direction of $x$ in $S$, so I can see the difference between the two.  Therefore, to find $T$, I think the first condition should be
$$
\begin{split}
\lVert Tx \rVert &= \sqrt{x^\top Q x} \\
\lVert Tx \rVert^2 &= x^\top Q x \\
(Tx)^\top Tx  &= x^\top Q x \\
x^\top T^\top Tx  &= x^\top Q x \implies T^\top T = Q
\end{split}
$$
But then I don't know how to continue. In particular, I don't know how to find the direction of $Tx$.   Where does it point to?  To me it seems that the problem is undetermined, and maybe the whole idea is misguided.
 A: -For simplicity, take the induced norm of the weighted inner product which is a quadratic function.
Let $Q_s$ and $Q_{as}$  be the symmetric and skewsymmetric part of Q, respectively. Then,
$x^TQx=x^T(Q_s+Q_{as})x=x^TQ_sx$
due to the fact that $x^TQ_{as}x=x^TQ^T_{as}x=-x^TQ_{as}x\implies x^TQ_{as}x=0$
What remains is the symmetric matrix that can be diagonalzed by the set of its eigen-vectors.
-Regarding your idea to find a transformation that preserves the magnitude and direction, this map is the identity only. And it is not helpful to change anything.
Instead of this you can search about a unitary transformation because it preserves the inner product value $i.e$. if U is unitary, then $<Ux,Uy>=<x,y>$.
In order to find a unitary transformation $U$ such that $U^{-1}QU=D$, Q must be normal matrix, where $D$ is diagonal.
A matrix $Q$ is said to be normal if $Q^TQ=QQ^T$.
A matrix $U$ is said to be unitary if $U^T=U^{-1}$
Regards,
A: Following up on Bilal's answer, the goal is to find a matrix $T$ such that
$$\text{d}_Q(Tx, Ty) = \text{d}(x, y)$$
where $d_Q$ is the "Q" distance function and $d$ is the standard (i.e. Euclidean) distance function.  (The original statement of the problem was wrong.)
So, we have
$$
\begin{split}
(Tx-Ty)^\top Q (Tx-Ty) &= (x-y)^\top (x-y) \\
(x-y)^\top T^\top Q T (x-y) &= (x-y)^\top (x-y) \\
\end{split}
$$
implying that $T^\top Q T = I$. Then it's clear that $T=Q^{-1/2}$. Since $Q$ is positive definite and thus orthogonally diagonalizable, we can find an orthogonal matrix $U$ and a diagonal matrix $S$ such that $Q=USU^\top$, and then we have $T=US^{-1/2}U^\top$.
For example, if
$$
Q=
\begin{bmatrix}
2 & -1 \\
-1 & 3/2
\end{bmatrix} \quad\text{then}\quad
T=
\begin{bmatrix}
.819 & .281 \\
.281 & .960
\end{bmatrix} .
$$
Now we can represent graphically a series of vectors $x_1, x_2,\dots$ and $Tx_1,Tx_2,\dots$ whose lengths in the Euclidean metric and in the metric $Q$ respectively coincide (and so do the respective inner products). In the plot below we can see some lines in black.  These lines are 1 unit of distance apart and form an orthogonal grid in the metric $Q$. For comparison see the same orthogonal grid in the Euclidean metric in light grey.

