Does $A^2 \geq B^2 > 0$ imply $ACA \geq BCB$ for square positive definite matrices?

Question

Assume we have two $n \times n$ real nondegenerate matrices $ A^2 $ and $B^2$, such that $$ A^2 \geq B^2 > 0, $$ where "$\geq$" means positive semidefinite (Loewner) ordering. Does the following inequality holds for any real matrix $C$ $$ ACA \geq BCB \ ? $$ If not, under which conditions on $C$ (or additional conditions on $A$ and $B$) does it holds?

I would appreciate any ideas, suggestions, counterexamples. Thanks!

Answer 1

I don't know if there are any nice (i.e. not-too-strong) conditions for the inequality to hold, but I'm sure that it doesn't always hold, even when $C$ is positive definite. Counterexample: \begin{align} A&=A^2=I,\\ B&=B^2=\frac12\pmatrix{1&1\\ 1&1},\\ C&=\operatorname{diag}(1,4). \end{align} In this case, we have $A^2\ge B^2\ge 0$ but $ACA-BCB=\frac14\pmatrix{-1&-5\\ -5&11}$ is indefinite. While $B$ is not positive definite here, by continuity, we can obtain a valid counterexample by adding a small positive multiple of $I$ to both $A$ and $B$.

Edit.

1. Note that if $ACA\ge BCB$ for all real symmetric $C$, we must have $A=B$ because $AIA\ge BIB$ and $A(-I)A\ge B(-I)B$ imply that $A^2=B^2$.
2. It isn't quite meaningful to consider $ACA\ge BCB$ for all $C\ge0$ either. In particular, if $A(vv^\ast)A\ge B(vv^\ast)B$ for every vector $v$, then $Bv$ must be equal to $\lambda_v Av$ for some $0\le\lambda_v\le1$. Therefore, by linearity, $B=\lambda A$ for some $0\le\lambda\le1$.
3. It is interesting to ask, however, if $A\ge B>0$ and $A^2\ge B^2$, what class of $C$ (under perhaps some additional conditions on $A$ and $B$) would satisfy the inequality $ACA\ge BCB$.

Answer 2

Here I posted a related question. My (preliminary) conclusion to this question might be a valuable idea/suggestion to your answer: I strongly assume (!) when your matrices commute (i.e. $(AB)^T AB = AB (AB)^T$) then $ACA \geq BCB$. This could be the case when $A$, $B$, and $C$ are diagonal matrices (and of course in the trivial cases of $C$ or $A$ and $B$ being multiples of $I$).