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As an excersise I'm trying to prove the following equality, where $a,b \in \mathbb{R}^n$. $$||a\cdot b^T||_2 = ||a||_2\cdot ||b||_2$$ I've made some progress, which I assume is correct ($\lambda_i$ is the ith eigenvalue of said matrix): $$\begin{align}||a\cdot b^T||_2 = \sqrt{\max_{i=1}^n\lambda_i((a\cdot b^T)^T\cdot a\cdot b^T)} = \sqrt{\max_{i=1}^n\lambda_i(b\cdot a^T\cdot a\cdot b^T)}\\=\sqrt{\max_{i=1}^n\lambda_i(||a||_2^2\cdot b\cdot b^T)} = ||a||_2\cdot\sqrt{\max_{i=1}^n\lambda_i(b\cdot b^T)}\end {align}$$ This is where I'm stuck. If $a=0$, then it's done. If not, then I should prove, that for any $b\in\mathbb{R}^n$: $$\sqrt{\max_{i=1}^n\lambda_i(b\cdot b^T)}=||b||_2$$ I'm kind of baffled by this problem and I'd really appreciate the help. Thanks!

Edit:

I've tried proving it an other way, but it's still no good, my approach in this attempt is from the fact that: $$||a\cdot b^T||_2 = \max_{x\neq0}\frac{||a\cdot b^T\cdot x||_2}{||x||_2}$$ Not showing the full calculations, but I got the following result: $$\max_{x\neq0}\frac{||a\cdot b^T\cdot x||_2}{||x||_2} = ||a||_2\cdot \max_{x\neq0}\frac{||b^T\cdot x||_2}{||x||_2}$$ Yet again, I'm stuck here. I should prove that: $$\max_{x\neq0}\frac{||b^T\cdot x||_2}{||x||_2} = ||b||_2$$

This approach seems more reasonable, but I still can't quite grasp how to finish the proof.

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You are almost done, there remains to remark that $\|b^T\cdot x\|_2\le \|b\|_2\cdot \|x\|_2$ by Cauchy-Schwartz inequality and the equality is attained for $x=\lambda b$, where $\lambda\in\Bbb R$.

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    $\begingroup$ I see! I haven’t thought about the Cauchy-Schwarz inequality, thanks! $\endgroup$ – Levente Kovács Nov 21 '17 at 10:43

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