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I have a matrix $Q\in\mathbb{R}^{m\times n}(m>n)$, where the columns of Q are unit and mutually orthogonal. $W\in \mathbb{R}^{m\times m}$ is a diagonal matrix with diagonal elements 0 or 1, and $WQ$ has full column rank (which means that there must be at least n diagonal elements equal to 1 in W). Now I'm trying to prove that the largest eigenvalue of $Q^TWQ$ is 1. I already know that $\Vert Q^TWQ\Vert_2\leq 1$, and if $\Vert Q^TWQ\Vert_2 < 1$, then for any $x\neq 0 \in \Bbb{R}^{n}$, $\Vert Q^TWQx\Vert_2\leq \Vert Q^TWQ \Vert_2 \Vert x\Vert_2 < \Vert x\Vert_2$. So, if there exists a vector $x\neq 0$, such that $\Vert Q^TWQx\Vert_2 = \Vert x\Vert_2$, it would imply that $\Vert Q^TWQ\Vert_2 = 1$. However, I have failed to find such a vector. Can anyone help? Or is this approach(conclusion) incorrect? Thanks a lot!

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    $\begingroup$ @TheoBendit You are right. $W$ is in $\mathbb{R}^{m\times m}$. Thank you very much! $\endgroup$
    – kai deng
    Commented Jul 10 at 12:07

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The conclusion is not correct. Take, for example: $$\pmatrix{\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}}\pmatrix{1&0\\0&0}\pmatrix{\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}}=\pmatrix{\frac{1}{2}},$$ which clearly has only the eigenvalue $\frac{1}{2}$.

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  • $\begingroup$ This example is nice! Can Q be extended to multi-columns? $\endgroup$
    – kai deng
    Commented Jul 10 at 12:34
  • $\begingroup$ @kaideng Certainly:$$\pmatrix{\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0&0\\0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}}\pmatrix{1&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0}\pmatrix{\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&0\\0&\frac{1}{\sqrt{2}}\\0&\frac{1}{\sqrt{2}}}=\pmatrix{\frac{1}{2}&0\\0&\frac{1}{2}}.$$Three columns is also available on request. :) $\endgroup$ Commented Jul 10 at 12:40
  • $\begingroup$ I get it, thank you so much! $\endgroup$
    – kai deng
    Commented Jul 10 at 12:52

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