Definitions
Consider the following optimization problem \begin{equation*}\arg \min_{x\in\mathbb{R}^n} \lVert y-x\rVert_{P}^2\end{equation*} where $y,P$ are given parameters and \begin{equation*} \lVert y-x\rVert_{P}^2 \triangleq (y-x)'P(y-x) \end{equation*} assume that $P$ is a regular covariance matrix, that is $P$ is symmetric and strictly positive definite ($P=P', P>0$).
Question
Under the previous assumption $P=P', P>0$, is it true that \begin{equation*}\arg \min_{x\in\mathbb{R}^n} \lVert y-x\rVert_{P}^2= \arg \min_{x\in\mathbb{R}^n} \lVert y-x\rVert^2 \end{equation*}?
Observation
In the simple case $2\times 2$, with the additional hypothesis that $P$ is diagonal, the equality seems to be true. Indeed, by denoting \begin{equation*} y=\left[\begin{array}{c} y_1 \\ y_2 \end{array}\right] \qquad x=\left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \qquad P=\left[\begin{array}{cc} P_{11} & 0\\ 0 & P_{22} \end{array}\right] \end{equation*} follows \begin{equation*} \lVert y-x\rVert_P^2= P_{11} (y_1-x_1)^2 + P_{22} (y_2-x_2)^2 \end{equation*} now, due to the sum sign the objective is to jointly minimize $P_{11} (y_1-x_1)^2$ and $P_{22} (y_2-x_2)^2$. These are two decoupled problems, so they can be treated separately, that is we are searching for \begin{equation*} \arg\min_{x_i \in \mathbb{R}} P_{ii} (y_i - x_i)^2 \qquad i=1,2 \end{equation*} now, since $P_{ii}$ is just a positive scaling factor, holds \begin{equation*} \arg\min_{x_i \in \mathbb{R}} P_{ii} (y_i - x_i)^2= \arg\min_{x_i \in \mathbb{R}} (y_i - x_i)^2 \qquad i=1,2 \end{equation*} and so rolling back to the original problem \begin{equation*}\arg \min_{x\in\mathbb{R}^n} \lVert y-x\rVert_{P}^2= \arg \min_{x\in\mathbb{R}^n} \lVert y-x\rVert^2 \end{equation*}
Remark
I've proved that the equality holds in the diagonal $2\times2$ case. The extension to the diagonal $n\times n$ case is identical. My difficulty is that I'm not sure how to treat the non-diagonal case.
$\nabla_x \left( \|y - x\|^2_P \right) = 2 P (y-x) = 0$, so we have $x = y$,
$$y = \arg\min_{x \in \mathbb{R}^n} \,\|y - x\|^2_P$$
$\nabla_x \left( \|y-x\|^2 \right) = 2(y-x) = 0$, so we have $x = y$,
$$y = \arg \min_{x \in \mathbb{R}^n} \, \|y-x\|^2$$