# Thread: Inequation with unit vectors

1. ## Inequation with unit vectors

Hi! I want to show that sum(norm(x_i-x_j), 1<=i<j<=n) <= n^2 for any vector space over R or C, where x_i are unit vectors.

Banach

2. There are ${n\choose 2}$ ways to select two indices $i. So
$\sum_{i

3. Hi! Thanks for your answer. I realized that i forgot to square the norm, so it should be shown: $\sum_{i and thus your idea does not work, since one has $4*{n \choose 2}$? How can this be solved?

4. Originally Posted by Banach
Hi! I want to show that $\sum_{1\leqslant i for any vector space over R or C, where x_i are unit vectors.
This is true if you are using a euclidean or Hilbert space norm, but it need not hold for other norms. For example, in the space $\mathbb{R}^2$ with the norm $\|(x,y)\| = |x|+|y|$, the four vectors $(\pm1,0),\ (0\pm1)$ are all distance 2 from each other, so the sum of the squares of the six distances is 24 which is greater than 16.

Here's how to prove the result for a real vector space where the norm comes from an inner product $\langle\,,\,\rangle$.

$\sum_{1\leqslant i

. . . . . . . . . . . . . . $= \sum_{i=1}^n\langle x_i,x_i\rangle - \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle$

. . . . . . . . . . . .. . $\leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle$

. . . . . . . . . . . .. . $= \sum_{i,j=1}^n\langle x_i,x_j\rangle = \Bigl\|\sum_{i=1}^nx_i\Bigr\|^2 \leqslant n^2$

The proof for complex spaces is more or less the same, with complex conjugates inserted where appropriate.

5. Hi, thanks for your answer. You are right, i forgot the prerequisite euclidian space.

Why does the inequation $\leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle
$
hold? If $\sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle
$
would be positive then it would be clear. But this is not the case. Can you explain this step to me please?

Greetings
Banach

6. $\sum_{1\leqslant i

. . . . . . . . . . . . . . $= \sum_{i=1}^n\langle x_i,x_i\rangle - \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle$

. . . . . . . . . . . .. . $=n-2\sum_{i

. . . . . . . . . . . .. . $\leq n+2\sum_{i by Cauchy-Schwarz
. . . . . . . . . . . .. . $n+2{n\choose 2}=n^2$

7. Originally Posted by Banach
Hi, thanks for your answer. You are right, i forgot the prerequisite euclidian space.

Why does the inequation $\leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle
$
hold? If $\sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle
$
would be positive then it would be clear. But this is not the case.
You're right, that was a careless mistake. See watchmath's correction above.

8. Alright, thank you both very much. Now everything is clear.

Another proof came to my mind:
One can show that $\sum_{1\leqslant i. Therefore the claim follows immediately.

Good night!

Banach