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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 $f(n)[t]$. How fast does $f(n)[t]$ grow with $n$? Does $f(n,t) = O(\log^{c}{n})$ when $t=O(n^{q})$ for some $c \in \mathbb{N}$ 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?

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    $\begingroup$ I am having parsing difficulties with this. Please rewrite it to be human-readable. $\endgroup$
    – Igor Rivin
    Commented Oct 11, 2011 at 18:34
  • $\begingroup$ Is this clear? My older version was when $t=1$. $\endgroup$
    – Turbo
    Commented Oct 11, 2011 at 19:09
  • $\begingroup$ Is $a_{ij1}$ a matrix for EACH i, or do you mean $(a_{ij})_{ij}$ is a matrix with entries $a_{ij}$ or...? I am confused... Maybe use two different letters instead of a third index to represent the pairs? $\endgroup$ Commented Oct 11, 2011 at 19:53
  • $\begingroup$ each aijk is a matrix $\endgroup$
    – Turbo
    Commented Oct 11, 2011 at 20:14

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