If $B \geq A \geq 0$ and $M \geq 0$ are real symmetric matrices, is it true that
$\| B M B \|_2 \geq \| A M A \|_2$ ?
2026-03-25 17:38:13.1774460293
Spectral norm inequality product
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No. It is known, for instance, that $B\ge A\ge0$ does not imply $B^2\ge A^2$. Now pick such a pair of real matrices $A$ and $B$. Then $u^T B^2u<u^T A^2u$ for some nonzero real vector $u$. Let $M=uu^T$. Then $$ \|BMB\|_2= u^TB^2u<u^TA^2u=\|AMA\|_2. $$