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Let matrices $X$ and $Y$ be positive semidefinite. Show that $X \succeq Y \succ 0$ is equivalent to $Y^{-1} \succeq X^{-1}$.

The teacher tells me that it is easy to prove via the Schur complement. But I could not get his idea. Could anyone help me and give me some intructions? Thanks in advance!

Rodrigo de Azevedo
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stander Qiu
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1 Answers1

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Hint. In my opinion, it is easier to prove that when $X,Y$ are positive definite, the following four statements are equivalent:

  • $X\succeq Y$,
  • $Y^{-1/2}XY^{-1/2}\succeq I$,
  • $\left(Y^{-1/2}XY^{-1/2}\right)^{-1}\preceq I$,
  • $X^{-1}\preceq Y^{-1}$.

Yet you can also follow your teacher's hint --- simply consider the Schur complements $M/Y^{-1}$ and $M/X$ for the matrix $$ M=\pmatrix{X&I\\ I&Y^{-1}}. $$

user1551
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  • Thanks for your help! I have already understood! – stander Qiu Oct 21 '17 at 06:49
  • Why is the matrix $M$ pd? Isn't the corresponding quadratic form only $v_1^T X v_1 + v_1^T v_2+ v_2^T v_1+ v_2^T Y^{-1} v_2$? We know by assumption that $v_1^T X v_1 > 0$ and $v_2^T Y^{-1} v_2 > 0$, but it's not clear to me how we can assume $v_1^T X v_1 + v_2^T Y^{-1} v_2 > -2v_1^T v_2$ for arbitrary $(v_1, v_2)$. And if we don't know that $M$ is pd, then we can't conclude that $M/X = Y^{-1} - X^{-1}$ is pd. – hasManyStupidQuestions Jan 28 '19 at 16:38
  • Never mind, by Wikipedia, $M$ is pd if and only if $Y^{-1}$ and $M/Y^{-1} = X$ are pd, and we know the latter to be true (since $Y$ pd implies $Y^{-1}$ pd). Which is why you said to consider both $M/Y^{-1}$ and $M/X$, not just $M/X$. – hasManyStupidQuestions Jan 28 '19 at 16:44
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    @hasManyStupidQuestions $M$ is congruent to both $X\oplus(M/X)$ and $(M/Y^{-1})\oplus Y^{-1}$. Therefore, given that both $X$ and $Y$ are positive definite, the statements $M/X\succeq0$ and $M/Y^{-1}\succeq0$ are equivalent. – user1551 Jan 28 '19 at 16:53
  • That is actually the clearest explanation of the psd/pd properties of Schur complement I've ever seen -- it actually makes sense to me now! Thanks so much for this. (For others who might initially be confused, $X \oplus (M/X)$ is the block matrix $[X & 0 \ 0 & M/X]$ and similarly $(M/Y^{-1}) \oplus Y^{-1}$ is the block matrix $[M/Y^{-1} & 0 \ 0 & Y]$. And the congruence transformation for $(M/Y^{-1}) \oplus Y^{-1}$ is $[I & 0 \ -Y & I]$ and for the congruence transformation for $X \oplus (M/X)$ is $[I & -X^{-1} \ 0 &I]$. – hasManyStupidQuestions Jan 28 '19 at 17:25