co.combinatorics - Orthogonal matrices with small entries

Here's an idea which I think might be expandable to a solution once some details are filled in. (I am rather tired at the moment, though, so apologies if there is a cretinous error in what follows.)

We'll do the case $n=4m-1$ where $m$ is an integer; the case $n=4m-3$ is similar.

Let $C$ be a $2mtimes 2m$ matrix which has the required form. Let $A$ be the $ntimes n$ matrix with $C$ in the top left corner, $1$ on the remaning $2m-1$ diagonal entries, and zero elsewhere. Let $B$ be the $ntimes n$ matrix with $C$ in the bottom right corner, $1$ on the remaining $2m-1$ diagonal entries, and zero elsewhere.

$A=left[begin{matrix} C & 0 \\ 0 & I_{2m-1} end{matrix} right]quad,quad B= left[begin{matrix} I_{2m-1} & 0 \\ 0 & C end{matrix} right]$

Both $A$ and $B$ will be real orthogonal since $C$ is.
Consider the matrix $AB$, which being the product of real orthogonal matrices will also be orthogonal. I claim that the entries will all be $O(sqrt{n})$ as required.

In more detail:

-- If both $i$ and $j$ are $leq 2m-1$, then $(AB)_{ij}=A_{ij}=C_{ij}$ which is small by our choice of $C$; by symmetry, we can dispose of the case where both $i$ and $j$ are $geq 2m+1$ in a similar way.

-- If $ileq 2m-1$ and $jgeq 2m+1$, then on considering $sum_r A_{ir}B_{rj}$ we see that the only nonzero contribution comes when $rleq 2m$ and $rgeq 2m$, i.e. when $r=2m$ and so $(AB)_{ij}=A_{i,2m}B_{2m,j}$ is small.

-- If $i=2m$ or $j=2m$ then a similar analysis shows that $(AB)_{ij}$ can't be bigger than the entries of $C$ (at least up to some constant independent of $m$).

-- If $igeq 2m+1$ and $jleq 2m-1$ then $(AB)_{ij}=0$.

That should handle the case $n=4m-1$. The case $n=4m-3$ can be done in a similar fashion, but this time we will have extra factors of $3$ floating around since we have $3times n$ and $ntimes 3$ regions to consider, rather than just $1times n$ and $ntimes 1$ regions.