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I'm trying to prove the following statement, which really feels like it should be correct.

If $A\succ B \succ 0$ for symmetric $A,B$, then show that $||Ax||_2 \ge || Bx||_2$ for vector $x$.

The closest I've gotten is using Cholesky factorizations and that $Ax = R^\top R x$ is 'similar' to $ x^\top R^\top R x $, but I can't progress any further. Is this statement even correct? If it is, then what about if $A$ and $B$ are not symmetric?

Edit: It turns about that this is not true for symmetric $A,B$, due to the comments below. So in this case, (this is very unlikely to be true, but)

If $A\succ B \succ 0$ for non-symmetric $A,B$, is $||Ax||_2 \ge || Bx||_2$ true?

Clarent
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  • This isn't true. See https://math.stackexchange.com/q/510895 – user1551 Dec 02 '22 at 18:18
  • @user1551 Could you help me understand why that post shows that the statement isn't true? I am struggling to understand. – Clarent Dec 02 '22 at 18:22
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    $|Ax|_2\ge|Bx|_2$ means that $x^TA^2x=|Ax|_2^2\ge|Bx|_2^2=x^TB^2x$. If this is true for all $x$, we must have $A^2\succeq B^2$. – user1551 Dec 02 '22 at 18:31
  • @user1551 I see! Then what if $A,B$ are not symmetric? I will edit the question! – Clarent Dec 02 '22 at 18:33
  • It doesn't make much sense to talk about $A\succ B$ when $A$ or $B$ aren't symmetric. At any rate, the statement remains untrue even if you allow $A$ and $B$ to be non-symmetric, because every symmetric matrix is the limit of a sequence of non-symmetric matrices. – user1551 Dec 02 '22 at 18:43

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