# show that $\det(D) = 0$ where $D_{ij} = |P_i - P_j|^2$

Let $$P_1,\cdots, P_6$$ be points in $$\mathbb{R}^3.$$ Let $$D$$ be the $$6\times 6$$ matrix whose $$(i,j)$$th-entry is the square of the distance between $$P_i$$ and $$P_j$$. Show that $$\det(D) = 0$$.

Note that it suffices to write $$D$$ as a sum of $$5$$ matrices of rank at most 1, since for any matrices A and B, $$\mathrm{rank}(A+B)\leq \mathrm{rank}(A) + \mathrm{rank}(B)$$ and so by induction $$\mathrm{rank}(\sum_{i=1}^n A_i) \leq \sum_{i=1}^n\mathrm{rank}(A_i)$$. Also $$D$$ is symmetric since $$|P_i - P_j| = |P_j-P_i|$$ for all $$i,j$$.

$$D$$ can trivially be written as the sum of $$6$$ matrices with rank at most 1; let a distinct row of D be the first row of each matrix while all other rows are zero. Suppose we have the specific example where $$P_1,P_2,\cdots, P_6 = (1,1,1),(1,1,2),(1,2,1),(2,1,1),(2,1,2),(2,2,1).$$ Then $$D = \begin{pmatrix}0 & 1 & 1 &2 & 2 & 2\\ 1 & 0 & 2 & 2 & 1 & 3\\ 1 & 2 & 0 & 2 & 3 & 1\\ 1 & 2 & 2 & 0 & 1 & 1 \\ 2 & 1 & 3 & 1 & 0 & 2\\ 2 & 3 & 1 & 1 & 2 & 0\end{pmatrix}.$$

One might be able to compute $$\det(D)$$ in this special case by using row or column operations. Also note that the determinant is $$n$$-linear and alternating.

If $$P_i=P_j$$ for some $$i\neq j$$ then $$D$$ will have two identical rows and hence determinant of zero, so we may suppose that all the $$P_i$$'s are distinct.

• What is your question/what do you need help with? Commented Oct 9, 2022 at 2:16
• The matrix $D$ that you show is not symmetric. $D_{1,4} \ne D_{4,1}$. Commented Oct 9, 2022 at 2:28
• hint: with $P:=\bigg[\begin{array}{c|c|c|c} \mathbf p_1 & \mathbf p_2 &\cdots & \mathbf p_{6} \end{array}\bigg]$ which has rank at most $3$, try writing $D$ as something involving $P$ plus 2 rank one matrices. Commented Oct 9, 2022 at 3:27

Let $$P$$ denote the matrix whose columns are $$P_1,\dots,P_6$$. Let $$e,v \in \Bbb R^6$$ denote vectors in $$\Bbb R^6$$ such that $$e = (1,\dots,1)$$ and $$v = (|P_1|^2,\dots,|P_6|^2)$$. Show that $$D = ev^T + ve^T - 2P^TP,$$ which is the sum of 2 rank-1 matrices and one rank-3 matrix.
• +1 I presume it should be $-2P^TP$? Commented Oct 9, 2022 at 16:59
• @BenGrossmann oh I see your idea. You basically just expanded $|P_i-P_j|^2 = |P_i|^2 + |P_j|^2 - 2 P_i\cdot P_j$ and you realized the connection to matrices, right? Commented Oct 11, 2022 at 16:14