Let $A, B\in \mathbb{R}^{n\times d}$ ($n \geq d$) be two matrices with $A^\top A \preceq B^\top B$. Let $D = \mathrm{diag}(d_1, \ldots, d_n)$ be a diagonal matrix with nonnegative entries. Do we still have $A^\top D A \preceq B^\top D B$?
My conjecture is not, but I wasn't able to come up with a simple counter-example. Any insights or hints would be much appreciated!
Just take $d=1.$ $a_1^2+\cdots+a_n^2\leq b_1^2+\cdots+b_n^2$ does not imply $d_1a_1^2+\cdots+d_na_n^2\leq d_1b_1^2+\cdots+d_nb_n^2.$