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Let $x=(x_1, \ldots, x_n)$ be a vector on $S^{n-1}$. Reorder coordinates such that $|x_1|\leq \ldots \leq |x_n|$.

I am wondering if there is a some relation between the absolute value of the coordinates of the vector $x$?

For example, to have all non-zero coordinates, I should have that, say, all of the coordinates would be $1/\sqrt n$. What happened if vector would have $k$ non-zero coordinates--would be the upper bound for $x_i$?

Thank you.

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  • $\begingroup$ But, $1/\sqrt n$ is not a lower or upper bound, its just the quadratic mean of the absolute values of the coordinates. $|x_1|\le 1/\sqrt n\le |x_n|$ is we can state. $\endgroup$
    – Berci
    Feb 13, 2013 at 0:01
  • $\begingroup$ @Berci: I did not get, why $|x_1|\leq 1/\sqrt n$? $\endgroup$
    – user62065
    Feb 13, 2013 at 0:07
  • $\begingroup$ We know $|x_1|\leq 1/\sqrt n$ because if not they would all be greater than $1/\sqrt n$ and the sum of squares would exceed $1$ $\endgroup$
    – Ross Millikan
    Feb 13, 2013 at 0:15

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Even if all the coordinates are non-zero, one could be almost $1$ and the rest could be very small. You can say that at least one coordinate is at least $\frac 1{\sqrt n}$ in absolute value, but that is about it.

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    $\begingroup$ (+1): We can also say that at least one coordinate is at most $\frac1{\sqrt{n}}$ in absolute value. $\endgroup$
    – Cameron Buie
    Feb 13, 2013 at 0:10
  • $\begingroup$ @CameronBuie: correct you are. Thanks. $\endgroup$
    – Ross Millikan
    Feb 13, 2013 at 0:14

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