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Let $A$ and $B$ be $n×n$ symmetric matrices such that $A,B\succeq0$ ($A$ and $B$ are positive semidefinite) and $A^2\succeq B^2$. Is the following inequality true? $$A\succeq B$$ In this answer, it's shown that (I couldn't understand the proof) if we replace "semidefinite" with "definite" in the above proposition, the inequality is true but I don't know whether it holds in this case.

S.H.W
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    The link you provide actually proves what you ask right ? What part of the proof don't you get ? – P. Quinton Nov 07 '23 at 10:52
  • @P.Quinton It proves if $A,B\succ 0$ and $A^2\succ B^2$ then $A\succ B$. I'm asking about the same proposition with $\succ$ replaced by $\succeq$. In the proof, it's assumed that $A^{-1}$ exists but a positive semidefinite matrix can be singular. – S.H.W Nov 07 '23 at 11:03
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    Well the third paragraph (the limiting argument) is about if $0\preceq B^2 \preceq A^2$ then $0\preceq B \preceq A$ when $0\preceq A,B$. – P. Quinton Nov 07 '23 at 14:21
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    if you want to do this directly without appealing to limits, consider: https://math.stackexchange.com/questions/3516718/if-0-le-a-le-b-then-sqrta-le-sqrtb/ – user8675309 Nov 07 '23 at 16:55
  • @user8675309 Thanks, I think it solves the problem. It would be great if you post it as an answer, so I can accept it. – S.H.W Nov 07 '23 at 18:23
  • @P.Quinton You're right, I didn't pay attention. By the way, I don't know how to make it rigorous. It holds for every $\epsilon \gt 0$ but we need some continuity condition to let $\epsilon \to 0$. – S.H.W Nov 07 '23 at 18:28
  • @S.H.W. With the definition of "positive semidefinite", you can show that the limit of a sequence of positive semidefinite matrices is positive semidefinite. – Ben Grossmann Nov 07 '23 at 18:43
  • @BenGrossmann That's really interesting. Would you provide a reference? – S.H.W Nov 07 '23 at 18:55
  • @S.H.W I could type a proof as an answer, but I don't have a reference offhand – Ben Grossmann Nov 07 '23 at 18:57
  • @BenGrossmann That's even better :). I would appreciate it. – S.H.W Nov 07 '23 at 19:01

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Per the "limiting argument" in the linked answer, it suffices to show that if $(A + tI) \succeq B$ for all $t > 0$, then $A \succeq B$.

Recall that a symmetric real matrix $M$ is positive semidefinite if for all vectors $x \in \Bbb R^n$, $x^TMx \geq 0$. Let $t_k = 1/k$. We note that for each $k$, $(A + t_k I) - B$ is positive semidefinite, so that $x^T[(A + t_k I) - B]x \geq 0$. Now, we have $$ x^T(A - B)x = \lim_{k \to \infty}x^T[(A + t_k I) - B]x. $$ The limit of a sequence of non-negative numbers is necessarily non-negative. Thus, $x^T(A - B)x \geq 0$. Since this holds for all vectors $x$, we conclude that $A - B \succeq 0$, which is to say that $A \succeq B$ as was desired.

S.H.W
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Ben Grossmann
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