Take the 2-minute tour ×

I'm trying to show that a vector norm $\|\cdot\|$ being absolute ($\|x\| = \|\;|x|\;\|)$ is equivalent to showing that $\|x'\| = \|[\alpha_1x_1\ldots\alpha_nx_n]^T\| = \|x\|$ for all $x\in\mathbb{C}^n$ and $|\alpha_i| = 1$ for all $\alpha_i$. I've shown that if $\|\cdot\|$ is absolute, then the given statement follows, but I'm having trouble showing the reverse.

From my proof of the first part, I know that $|x'| = |x|$, so I can either show that $\|x'\| = \|\;|x'|\;\|$ or take the direct route of showing that $\|x'\| = \|\;|x|\;\|$. Either way, I don't see how to proceed. Intuition tells me that the crucial step will revolve around using the fact that $|\alpha_i| = 1$, so that's where I've started, but no luck so for. I'll update if I find anything more out, but a nudge in the right direction would be much appreciated.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let $x\in\mathbb{C}$ be arbitrary and specialize (pick out) each $\alpha_i$ individually such that $\alpha_ix_i=|x_i|$ and the modulus is unity ($|\alpha_i|=1$) for each $i$ - the given hypothesis then implies $\|x'\|=\|[\alpha_ix_i]^T\|=\||x|\|.$

share|improve this answer
Doesn't this violate the hypothesis that this must hold for all $\alpha_i$? I read "...for all $x$ and all $y$..." to mean any combination of arbitrary $x$ and $y$ –  brc Nov 1 '11 at 6:39
@brc: You get $\|x'\|=\|x\|$ by hypothesis. Also, since it holds for any $|\alpha_i|=1$, you're free to make them whatever you want in order to continue investigating - it changes nothing. By choosing them so $\alpha_ix_i=|x_i|$ (see how this is possible) you can check directly that $\|x'\|=\||x|\|$. Hence $\|x\|=\||x|\|$. –  anon Nov 1 '11 at 6:43
That makes sense. The explanation is/was clear, I was just worried about the assumptions made to derive it. Thanks. –  brc Nov 1 '11 at 6:46

Your Answer


