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For real symmetric matrices $K$ and $\hat{K}$, if $u^{T}{K}u\le u^{T}\hat{K}u\le C u^{T}{K}u$ for all $u$ in the row space of $K$, then $u^{T}{K^{+}}u\ge u^{T}\hat{K}^{+}u\ge u^{T}{K^{+}}u/C$ for all $u$ in the row space of $K$, where $K^{+}$ is the Pseudoinverse of $K$.

I found this claim in the paper http://arxiv.org/abs/1407.1289. Is there a simple proof for this?

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    $\begingroup$ I assume that $K$ and $\hat{K}$ are supposed to be at least hermitian? This surely isn't true for all matrices. $\endgroup$
    – Nathaniel Johnston
    Commented Dec 7, 2014 at 17:34
  • $\begingroup$ yes they are real symmetric $\endgroup$
    – user29261
    Commented Dec 7, 2014 at 17:40

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