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This question proves that $||QA||_2 = ||A||_2$:

$||QA||_2 = \sup_{||x||=1}{||QAx||}=\sup_{||x||=1}{\sqrt{(QAx)^T(QAx)}} = \sup_{||x||=1}{\sqrt{x^TA^TQ^TQAx|}} = sup_{||x||=1}{\sqrt{x^TA^TAx}} = ||A||_2$

How to prove it for $||AQ||_2$ too?

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  • $\begingroup$ You could use your approach together with $\|B^T\|_2 = \|B\|_2$ $\endgroup$ – Wauzl Oct 17 '17 at 10:52
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We denote $\langle x, y\rangle_2 = x^Ty$. Then you have $$ \Vert Qx \Vert_2 = \langle Qx, Qx\rangle_2 = \langle x, Q^TQx\rangle_2 = \langle x, x\rangle_2 = \Vert x \Vert_2.$$ Therefore we have $\{x \in \mathbb R^n : \Vert x \Vert_2 = 1\} = \{x \in \mathbb R^n : \Vert Qx \Vert_2 = 1\}$. Thus we get $\sup_{\Vert x \Vert_2 = 1}{\Vert AQx \Vert_2}= \sup_{\Vert Qx \Vert_2 = 1}{\Vert A Qx \Vert_2}$ for the suprema.

Hence you get $$\Vert AQ \Vert_2 = \sup_{\Vert x \Vert_2 = 1}{\Vert AQx \Vert_2}= \sup_{\Vert Qx \Vert_2 = 1}{\Vert A Qx \Vert_2} = \sup_{\Vert y \Vert_2 = 1}{\Vert Ay \Vert_2} = \Vert A \Vert_2.$$

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    $\begingroup$ Could you explain why it's okay to substitute x with Qx? I understand they have the same norm but I don't see implication ||Qx||=||x|| => ||AQx||=||Ax|| $\endgroup$ – Tomasz Garbus Oct 17 '17 at 10:59
  • $\begingroup$ I edited my answer accordingly to make it more clear :) $\endgroup$ – Yaddle Oct 17 '17 at 11:16
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The map $Q$ preserves the unit sphere, so the sup of $\|AQx\|$ over the unit sphere will be equal to the sup of $\|A\|$ over the unit sphere.

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