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It may be better to post this question here. Assume that $M$ is an $m$ by $m$ ($m$ is an even number) symmetric positive-semi-definite matrix with exactly $m/2$ positive eigenvalues and every entry of $M$ is between $0$ and $1$. Let $I_{r}$ be an $m$ by $m$ diagonal matrix with $m/2$ of $1$'s on its diagonal and other diagonal entries being zeros (different arrangements of $1$'s and zeros on the diagonal give different $r$). Would $m^{2}/\inf_M\sup_{r}\left(\left\Vert \left(I_{r}MI_{r}\right)^{\dagger}\right\Vert \right)$ be bounded? where $X^{\dagger}$ is the extended pseudoinverse of $X$, defined as $\left(U\mbox{diag}\left(d_{1},\ldots,d_{m/2},0,\ldots,0\right)U^{T}\right)^{\dagger}=U\mbox{diag}\left(d_{1}^{-1},\ldots,d_{m/2}^{-1},0,\ldots,0\right)U^{T}$ in which $d_{i}^{-1}=\infty$ if $d_{i}=0$. Thanks.

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Better to post it here rather than where? –  David Roberts Oct 29 '12 at 6:46
    
David, rather than se. –  io0 Oct 29 '12 at 8:06
    
Bounded above as a function of $m$? But you could take $M = I_r$, so the denominator is at most $1$, and your expression is at least $m^2$. –  Robert Israel Oct 29 '12 at 16:39
    
Robert, yes, it means whether it is bounded above as a function of $m$. You are right. Sorry, the definition of the inverse in the question needed to be changed to the extended inverse. –  io0 Oct 29 '12 at 17:41
    
Now I really don't understand. $I_r M I_r$ has rank at most $r$, so what could $\|(I_r M I_r)^\dagger\|$ possibly be other than $\infty$? –  Robert Israel Oct 29 '12 at 22:14

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