Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
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|cite|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


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.