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!


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.


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 $\lambda\in\Bbb R$.

  • 1
    $\begingroup$ 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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