# l2-norm of matrix products

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!

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

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 $$\frac{1}{2}$$.
• @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. :) Commented Jul 10 at 12:40