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Let $A, B$ be two positive semidefinite matrices. Prove that if both $A-B$ and $B-A$ are positive semidefinite, then $A = B$.

I can show that their diagonal elements are the same but for others I have no idea. Any help will be appreciated.

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One route is to observe that if $\lambda,v$ is an eigenvalue/eigenvector pair of $M$ with , then $(-M)v=-\lambda v$, meaning that $-\lambda$ is an eigenvalue of $(-M)$. Since $M:=A-B$ is positive semidefinite, conclude that any eigenvalue of $M$ satisfies $0\leq \lambda\leq 0$, so $\lambda=0$, so that $M=0$.

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  • $\begingroup$ Thank you very much! Thats a neat proof $\endgroup$ – honzaik Mar 15 at 18:15
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By assumption $A-B=-(B-A)$ is both positive semidefinite and negative semidefinite. This can only happen if $A-B=0$.

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  • $\begingroup$ its this equivalent to saying that $x^TAx=0 \forall x \implies A = 0$ which is wrong (math.stackexchange.com/questions/358281/…) ? $\endgroup$ – honzaik Mar 15 at 17:57
  • $\begingroup$ I assumed we are working over $\mathbb C$ and so we are using $x^*Ax$. So the link is not relevant. $\endgroup$ – chhro Mar 15 at 18:01
  • $\begingroup$ actually, the argument still works since the real positive semidefinite matrices=real symmetric matrices with nonnegative eigenvalues. I don't see the point of mentioning the link. I didn't use that in the argument. $\endgroup$ – chhro Mar 15 at 18:05
  • $\begingroup$ well i assumed you meant $0 \geq -x^T (B-A) x = x^T (A-B) x \geq 0$. I guess when I hear positive semidef etc I assume this definition with vectors. $\endgroup$ – honzaik Mar 15 at 18:10

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