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Let $A,B,C$ be real symmetric positive definite matrices. For some small $\epsilon \ll 1$, suppose $(1-\epsilon)BB^{\top}\preceq AA^{\top} \preceq (1+\epsilon)BB^{\top}$, do we have $(1-\epsilon)BCB^{\top}\preceq ACA^{\top} \preceq (1+\epsilon)BCB^{\top}$?

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No. Let $B=I$ and $A=\pmatrix{a&b\\ b&c}$ be any positive definite matrix such that $b\ne0$ and $(1-\epsilon)I\preceq A^2\preceq(1+\epsilon)I$. Let $C'=\operatorname{diag}(1,0)$. Then $$ AC'A=\pmatrix{a\\ b}\pmatrix{a&b}=\pmatrix{a^2&ab\\ ab&b^2}, \ (1+\epsilon)BC'B=\pmatrix{1+\epsilon&0\\ 0&0}. $$ Thus the inequality $AC'A\preceq(1+\epsilon)BC'B$ does not hold, because the bottom right entry of the LHS (i.e. $b^2$) is greater than the corresponding entry of the RHS (namely, $0$). Therefore it also doesn't hold when $C$ is any positive definite matrix that is sufficiently close to $C'$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Zhao Song Apr 6 at 6:53

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