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Thanks in advance.

I'm struggling with some statement in a paper. It claims as follows,

For a p.s.d. operator $H:\mathbb{R}^d\to\mathbb{R}^d$ with an inverse and some $v\in\mathbb{R}^d$, such that

$v^{T}Hv\le\gamma$

for some constant $\gamma > 0$,

then $vv^T\preceq \gamma{H^{-1}}$

I cannot see it is obvious. Please help.

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migrated from mathoverflow.net Apr 2 '18 at 9:33

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  • $\begingroup$ Consider $1 \times 1$ matrices: $H=-1$, $v=\gamma=1$. Then we seem to be saying that if $-1\le 1$ then $1 \le -1$. Check again to see if you have the correct statement. $\endgroup$ – Ben McKay Mar 31 '18 at 6:40
  • $\begingroup$ The proof below assumes that $H$ has a symmetric square root, i.e. that $H$ is positive definite. It is clearly wrong otherwise, as my example proves. $\endgroup$ – Ben McKay Mar 31 '18 at 6:48
  • $\begingroup$ Ohhh. I forgot it ... Really thx. $\endgroup$ – Morino_Hikari Mar 31 '18 at 6:49
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For any $x\in\mathbb{R}^d$, we have the following

$x'vv'x=v'xx'v=v'H^{1/2}H^{-1/2}xx'H^{-1/2}H^{1/2}v$

$\le trace(H^{-1/2}xx'H^{-1/2})v'Hv=x'H^{-1}x*v'Hv\le\gamma x'H^{-1}x$

The conclusion follows by the arbitrariness of $x$.

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  • $\begingroup$ Thanks for your answer^^ And the proof will be correct with an additional condition that H is positive definite. [That's true there is such a property of the operator considered in that paper. I forgot it in my post. Again thanks Ben's hints as well.] $\endgroup$ – Morino_Hikari Mar 31 '18 at 7:04

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