$AA^{t}=A^{t}A$ 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, $0&lt;c&lt;d$ (note that this is a strict inequality). That is, $M$ look like this:

$\begin{pmatrix}d &amp; c &amp; \ldots &amp; \ldots &amp; c &amp; c\\c &amp; d &amp; \ldots &amp; \ldots &amp; c &amp; c\\\ldots &amp; \ldots &amp; \ldots &amp; \ldots &amp; \ldots &amp; \ldots\\\ldots &amp; \ldots &amp; \ldots &amp; \ldots &amp; \ldots &amp; \ldots\\c &amp; c &amp; \ldots &amp; \ldots &amp; d &amp; c\\c &amp; c &amp; \ldots &amp; \ldots &amp; c &amp; d\\\end{pmatrix}$

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.
Here's a two-dimensional counterexample:

$$A = \left[\begin{array}{cc}0 &amp; 5\\3 &amp; 4\end{array}\right].$$
Even for the case $c=d$ it can be valid, as

$$
\textbf{A} = \left[
\begin{array}{cccc}
\sqrt{c/n}&amp;\sqrt{c/n}&amp;\cdots&amp;\sqrt{c/n}\\
\sqrt{c/n}&amp;\sqrt{c/n}&amp;\cdots&amp;\sqrt{c/n}\\
\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\
\sqrt{c/n}&amp;\sqrt{c/n}&amp;\cdots&amp;\sqrt{c/n}\\
\end{array}
\right],
$$

has the property that

$$
\textbf{A} \textbf{A}^T = \textbf{A}^T \textbf{A} = \textbf{M},
$$

where

$$
\textbf{M} = \left[
\begin{array}{cccc}
c&amp;c&amp;\cdots&amp;c\\
c&amp;c&amp;\cdots&amp;c\\
\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\
c&amp;c&amp;\cdots&amp;c\\
\end{array}
\right],
$$