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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.

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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|\|.

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    That makes sense. The explanation is/was clear, I was just worried about the assumptions made to derive it. Thanks.2011-11-01