Suppose you are given all $\ell_p$ norms of a vector $v\in \mathbb R^d$. Is it true that the set of all its $\ell_p$ norms $\{\|v\|_{p},p=1,..,\infty\}$ uniquely define the vector $v$ up to permutations of its entries?
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1 Answer
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3
Hint.
Yes this is true up to a permutation of the absolute values of the coordinates. This is a consequence of Newton's identities: if you know the $\Vert \cdot \Vert_p$, you know the elementary symmetric polynomials. Hence the initial vector up to permutations.
answered Aug 27, 2015 at 19:10
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$\begingroup$ Thanks for your answer. I was looking at the link you sent but is still not clear to me. Could you please be more precise about "Hence the initial vector up to permutations."? Thanks! $\endgroup$– FabioCommented Aug 28, 2015 at 1:22
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$\begingroup$ how do you conclude the signs of $v$'s components using the $\ell_p$ norm alone? $\endgroup$ Commented Aug 28, 2015 at 1:55
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$\begingroup$ I updated the answer as indeed the $\ell_p$ Involves the absolute values of the coordinates and not the coordinates themselves directly. $\endgroup$ Commented Aug 28, 2015 at 5:46