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I want to determine whether there exists $5$ points in $\mathbb R^4$ such that the following matrix is the distance matrix. $$ \begin{pmatrix} 0& \sqrt5 & \sqrt5 & \sqrt5 & \sqrt5 \\ \sqrt5 & 0& 2\sqrt5 & 2\sqrt2 & 2\\ \sqrt5 &2\sqrt5 & 0 & 2 & 2\sqrt2 \\ \sqrt5 & 2\sqrt2 & 2 & 0 &2 \sqrt5 &\\ \sqrt5 & 2 & 2\sqrt2 & 2 \sqrt5 &0\\ \end{pmatrix}.$$ I tried to assume that $x_0=(0,0,0,0)$, $x_1=(x_{11},0,0,0)$, $x_2=(x_{21}, x_{22}, 0,0)$, etc.. but I got stuck because I got $x_{22}=0$

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marked as duplicate by joriki, Nate Eldredge, Stefan Hansen, Davide Giraudo, Sasha Feb 24 '13 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Not a solution but perhaps a simplification. Notice that all points lie on the sphere of radius $\sqrt{5}$ around point $1$. Points $3$ and $2$ have distance $2\sqrt{5}$ so they are antipodal on the sphere. Same with points $4$ and $5$. This means that the points lie on two lines intersecting at point $1$. – EuYu Feb 24 '13 at 14:43
up vote 2 down vote accepted

This is impossible.

So take $x_1$ and observe that the four other points are on the sphere of radius $\sqrt{5}$ centered at $x_1$.

Then note that $$ \|x_2-x_3\|=\|x_4-x_5\|=2\sqrt{5}. $$ So $x_2$ (resp $x_4$) is the symmetric of $x_3$ (resp $x_5$) with respect to $x_1$.

In particular, you should have ($[x_4,x_5]$ is diameter of the sphere, Pythagoras applies) $$ \|x_4-x_5\|^2=\|x_4-x_2\|^2+\|x_2-x_5\|^2. $$

But this yields $20=8+4$. So such points don't exist.

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Thank you very much! – user2102173 Feb 24 '13 at 16:19

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