Consider given pairs of variables: $a_{ir1},a_{ir2}\in \mathbb{R}^{m \times m}$ and $a_{jr1},a_{jr2}\in \mathbb{R}^{m \times m}$, where $r \in \{1,2,\cdots,t\}$, consider the constraints:
$\sum_{r=1}^{t}[a_{ir1}a_{jr2}+a_{ir2}a_{jr1}]=\sum_{r=1}^{t}[a_{jr1}a_{ir2}+a_{jr2}a_{ir1}]=0$
$\sum_{r=1}^{t}[a_{ir1}a_{ir2}+a_{ir2}a_{ir1}]=\sum_{r=1}^{t}[a_{jr1}a_{jr2}+a_{jr2}a_{jr1}]=I$
with $i \ne j$ and $i,j \in \{1,2,\cdots,n\}$.
Let the smallest size of matrices such constraints as a function of $n$ and $t$ be $f(n,t)$. My primary question is how fast does $f(n,t)$ grow with $n$ and $t$? For a fixed $t$, let the growth be $f(n)$. f(n)[t]$. How fast does $f(n)$ f(n)[t]$ grow with $n$? Does $f(n) f(n,t) = O(\log^{c}{n})$ when $t=O(n^{q})$ for some $c \in \mathbb{N}$?mathbb{N}$ and $\frac{1}{3} > q \in \mathbb{Q}$?
Secondly, how do you find such matrix solutions explicitly?
[Note: Each $a_{ijk}$ is a square matrix.]
What I know: For $t=1$, I am fairly certain $f(n,1) = n$. For any fixed $t$, I don't think we can do better (although not sure). What happens when $t$ is allowed to grow although with $n$ although at a sub-cubic rate w.r.t $n$ is something I am interested?
