Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 less than $1$. 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. Would $m^{2}/\inf_M\sup_{r}\left(\left\Vert \left(I_{r}MI_{r}\right)^{\dagger}\right\Vert \right)$ be always 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.

share|cite|improve this question
Did you mean $\sup_M$? The question makes no sense in its current form; you introduced $M$ as a given $m\times m$ matrix, and now you want to take the limit $m\to\infty$, but you don't specify how $M$ varies with $m$. The $r$ subscript on the $\sup$ also doesn't make sense because $r$ isn't actually a variable that $I_r$ depends on. – joriki Oct 29 '12 at 0:45
joriki, you are right. But it means that when M satisfies the assumption, one considers all the permutations of $m/2$ of $1$'s and $m/2$ of $0$'s, which gives all possible $I_r$. I should put it as whether it is bounded. – Userma Oct 29 '12 at 4:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.