In a quadratic program (QP), do linear equality constraints always reduce the norm of the minimizer? Specifically, let $P \succ 0$, $A \in \mathsf{M}_{m\times n}$ and $q\in\mathbb{R}^n$. Define

$$x^* := \arg\min_x\,\tfrac{1}{2} x^\mathsf{T} P x - q^\mathsf{T}x$$

and

\begin{align} x_c^* &:= \arg\min_x \, \tfrac{1}{2} x^\mathsf{T} P x - q^\mathsf{T}x\\ &\quad\,\,\,\operatorname{subject to} \,\,Ax=0. \end{align}

Intuitively $\|x_c^*\| \leq \|x^*\|$, if not in the standard $\ell^2$ norm in the $P$ (or maybe $P^{-1}$) induced norm $\|x\|_P = \langle Px,x \rangle^{1/2}$, because I'd think that the solution $x_c^*$ is the $\|\cdot\|_P$ metric projection of $x^*$ onto $\ker A$, a closed convex set, and such a projection is a contraction.

Nevertheless, I'm having trouble showing this. Boyd and Vandenberghe [p.546] tell us $x_c^* = (I + P^{-1}A^\mathsf{T}(AP^{-1}A^\mathsf{T})^{-1}A)P^{-1}q$ while $x^* = P^{-1}q$. Hence it suffices to show the operator $I + P^{-1}A^\mathsf{T}(AP^{-1}A^\mathsf{T})^{-1}A$ is a contraction under some metric.

Unfortunately, I just sampled a random $A$ and positive $P$, and the above operator is not a contraction in the $\ell^2$-norm in general.

Questions:

- is $\|x_c^*\|_2 \leq \|x^*\|_2$ in general?
- if not, is this true under a different norm such as $\|\cdot\|_P$?

If possible, a bound not involving $A$ would be helpful.