This is a a small $n$ restriction of another question.
Find a $5\times 5$ matrix of unit vectors $\xi_{ij}\in \mathbb{C}^5$ such that:
- Entries along rows and columns are orthogonal, that is for $1\leq i,j,k,l\leq 5$: $$\delta_{i,k}+\delta_{j,l}=1\implies \langle \xi_{ij},\xi_{kl}\rangle=0.$$
- Entries not on a common row or column are neither parallel, anti-parallel, nor orthogonal, again for $1\leq i,j,k,l\leq 5$:: $$\delta_{i,k}+\delta_{j,l}=0\implies 0<|\langle \xi_{ij},\xi_{kl}\rangle|<1$$
This is proving a real wicked problem. Motivation and an $n=4$ example in the original questions.
Attempts thus far:
I have started with the first two rows :
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} \\ \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} \end{bmatrix}$$
To get orthogonality along the first row, we are multiplying four numbers, and along columns, just three. I started with 24th roots of unity but then this reduced to sixth roots, powers of $w=\exp(2\pi i/6)$. You can get some of what you want with this by considering what combinations of four and three sixth roots give zero. It seems possible to get all but one of:
- orthogonal along rows one, two, three
- appropriate orthogonality between rows one and two, and one and three
- appropriate orthogonality between rows two and three
But it seems to fail before it ever gets to rows four or five, or indeed the second condition of being non-orthogonal nor parallel, nor anti-parallel. This wasn't the best I did, but something like:
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1\\ 1\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ -1\\ -1\end{pmatrix} & \begin{pmatrix} 0 \\ w \\ w^4 \\ 1\\ -1\end{pmatrix} & \begin{pmatrix} 0 \\ 1\\ -1 \\ w^2\\ w^5\end{pmatrix} \\ \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^2\\ w^4\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^5\\ 1\end{pmatrix} & \begin{pmatrix} w^2 \\ 0 \\ w^4 \\ w^2\\ w\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ -1 \\ w^4\\ -1\end{pmatrix} \\ \begin{pmatrix} 0 \\ 0 \\ 2 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 1\\ w \\ 0 \\ w^5\\ -1\end{pmatrix} & \begin{pmatrix} w^2 \\ w \\ 0 \\ w^2\\ 1\end{pmatrix} & \begin{pmatrix} -1\\ w^2 \\ 0 \\ w^5\\ 1\end{pmatrix} & \begin{pmatrix} w^2 \\ w \\ 0 \\ w\\ w^2\end{pmatrix} \end{bmatrix}$$
This attempt falls down on orthogonality in row three, and orthogonality in row 2 vs row 3.
