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Given two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ on a finite-dimensional vector space, could it be that there exist vectors $\vec{v}_1,\vec{v}_2$ such that $$\|\vec{v}_1\|_a>\|\vec{v}_2\|_a$$ and $$\|\vec{v}_1\|_b<\|\vec{v}_2\|_b$$ i.e. that the norms "disagree" on which vector is the longest?

Maybe this is related to norm equivalence, but I don't quite see how to get here from there.

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    $\begingroup$ A vector's $\ell_\infty$-norm can be large even if the vector has only one nonzero component. So our two vectors could be $(1,1,1,1,1)$ and $(0,0,0,0,1.01)$, for example. $\endgroup$ – littleO Sep 30 '16 at 12:33
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Yes. Otherwise the norms would only be multiples of each other.

$$\|(30,40)\|_2=50 > 45 =\|(0,45)\|_2$$ $$\|(30,40)\|_\infty=40 < 45 =\|(0,45)\|_\infty$$

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Sure. An example in $\Bbb{R}^2$: \begin{align} ||(x,y)||_a &= \sqrt{\left( \frac{x}{2} \right)^2 + y^2} \\ ||(x,y)||_b &= \sqrt{x^2 + \left( \frac{y}{2} \right)^2} \end{align} The sets of constant norm under these two norms are ellipses, but the major axes with the $a$ norm are along the $x$-axis and the major axes for the $b$-norm are long the $y$-axis. This suggests looking at \begin{align} ||(0,1)||_a &= 1 & ||(1,0)||_a &= 1/2 \\ ||(0,1)||_b &= 1/2 & ||(1,0)||_b &= 1 \text{.} \end{align}

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