$AA^{t}=A^{t}A$ in this case? Let $M$ be a square matrix such that $M_{i,j}=c$ if $i\not=j$ and $d$ if $i=j$. Also, $0<c<d$ (note that this is a strict inequality). That is, $M$ look like this:
$\begin{pmatrix}d & c & \ldots & \ldots & c & c\\c & d & \ldots & \ldots & c & c\\\ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\\ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\c & c & \ldots & \ldots & d & c\\c & c & \ldots & \ldots & c & d\\\end{pmatrix}$
If $A$ is a square matrix such that $AA^{t}=M$ is it necessarily the case that $A^{t}A=M$?
(we are working in $\mathbb{R}$; actually the actual question is meant for $\mathbb{Q}$ but I am trying to gain some insight by looking at the easier situation)
This is clearly true if we allow $c=0$, and is clearly false if we allow $c=d$. But I have not been able to do it in this case. I suspect it is false but have not been able to find a counterexample.
 A: Here's a two-dimensional counterexample:
$$A = \left[\begin{array}{cc}0 & 5\\3 & 4\end{array}\right].$$
A: I think perhaps this serves as a counterexample.
If $c < \frac{d}{n-1}$, then the matrix is Hermitian and strictly diagonally dominant with positive entries, and hence is positive semi-definite. Therefore, it admits a unique Cholesky factorization, $U^TU = M$ (equivalently $LL^T = M$). However, it is not necessarily true that $UU^T = M$. For example, let $$M = \begin{pmatrix} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 &3 \end{pmatrix}.$$
Then, $$U \approx \begin{pmatrix}1.7321 &   0.5774   & 0.5774\\
         0    & 1.6330    & 0.4082 \\
         0         & 0    & 1.5811 \end{pmatrix}$$
and it can be seen that $U^TU = M$ but not $UU^T$.
A: Even for the case $c=d$ it can be valid, as
$$
\textbf{A} = \left[
\begin{array}{cccc}
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\vdots&\vdots&\ddots&\vdots\\
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\end{array}
\right],
$$
has the property that
$$
\textbf{A} \textbf{A}^T = \textbf{A}^T \textbf{A} = \textbf{M},
$$
where
$$
\textbf{M} = \left[
\begin{array}{cccc}
c&c&\cdots&c\\
c&c&\cdots&c\\
\vdots&\vdots&\ddots&\vdots\\
c&c&\cdots&c\\
\end{array}
\right],
$$
