matrix identity extends to real entries? Let a matrix be $(m+n)$ by $(m+n)$ be all reals with determinant $1.$ Call the matrix $S.$ Name the top $m$ rows $A$ and the next $n$ rows $V.$ Now, $S$ has an inverse $T,$  $ST = I.$ Name the left $m$ columns $W^T$ and the right $n$ columns $B^T.$
We get two Gram matrices, first $G = A A^T$ is $m$ by $m.$ Next, $H = B B^T$ is $n$ by $n.$
All the examples I have done so far have
$$\det(G) = \det(H)$$
Theorems:
(I) true when integer entries
(II) true when $S \in SO_n$
(III) true for 2 by 2, that would be $m=n=1$
The question is, is this thing always true with real entries? i have a limited ability to look for counterexamples in low dimension, none so far. I have tried a few examples up to $m+n \leq 4,$ however rational entries with small denominators. I also cannot prove it, it is in the area of Schur complement but I do not see a proof either. I should add that, should the thing be true over the reals, i would expect a proof to have appeared in the (few) places where there is a proof with integer entries.
I did one by hand last night, irrational entries. Let $w = 2 \cos \frac{2 \pi}{9},$ so that $w > 1$ and $w^3 - 3w + 1 = 0.$ The initial matrix with determinant $1$ is
$$
S =
\left(
\begin{array}{ccc}
w & 1 & 1 \\
1 & w & 1 \\
1 & 1 & w \\
\end{array}
\right)
$$
with 
$$
T =
\left(
\begin{array}{ccc}
w^2-1 & 1-w & 1-w \\
1 -w & w^2-1 & 1-w \\
1-w & 1-w & w^2-1 \\
\end{array}
\right)
$$
From $$ A = ( w,1,1)$$
we get $G = \left( w^2 + 2 \right)$ and
$$ \det G = w^2 + 2. $$
Then
$$
B =
\left(
\begin{array}{ccc}
1 -w & w^2-1 & 1-w \\
1-w & 1-w & w^2-1 \\
\end{array}
\right)
$$
gives, using relations $w^3 = 3 w - 1$ and $w^4 = 3 w^2 - w,$
$$
H =
\left(
\begin{array}{cc}
3 w^2 - 5 w + 3 & 3w^2-6w+1  \\
3w^2 - 6 w + 1 & 3 w^2 - 5 w + 3  \\
\end{array}
\right)
$$
and, after using the same relations,
$$ \det H = w^2 + 2 = \det G $$

 A: Example with integer coefficients I had partly typed up. Detail that could be important, could be irrelevant: here, i did not work out a complete matrix $S$ that contained $A$ as the top two rows. i went directly to putting $AR$ in Hermite form by column operations. Hmmm: with the square matrix $R$ far below, the top two rows of $R^{-1}$ really do turn out to be $A.$ maybe this is not a problem.
The whole point of the exercise is that the determinant of the Gram matrix  for $M$ is the same as 
the determinant of the Gram matrix  for $M^\perp \; .$
We begin with a lattice $M$ with basis given by  the two rows of
$$
A =
\left(
\begin{array}{ccccc}
1&1&0&0&2 \\
1&1&1&0&3
\end{array}
\right)
$$
The inner products give the Gram matrix $G = A A^T$ as
$$
G =
\left(
\begin{array}{cc}
6&8 \\
8&12
\end{array}
\right)
$$
This gives
$$
G^{-1} =
\left(
\begin{array}{cc}
\frac{3}{2}&-1 \\
-1&\frac{3}{2}
\end{array}
\right)
$$
This allows us to define a basis with rational elements, for the dual lattice $M^\ast,$ given by $A^\ast = G^{-1}A.$ The dual lattice is integer valued linear functionals on $M,$ but we can expres it using rational vectors. 
$$
A^\ast =
\left(
\begin{array}{ccccc}
\frac{1}{2}&\frac{1}{2}&-1&0&0 \\
-\frac{1}{4}&-\frac{1}{4}&\frac{3}{4}&0&\frac{1}{4}
\end{array}
\right)
$$
We have, automatically, $A^\ast A^T = I$
We also need the Smith Normal Form of $G,$ in symbols $USV = G.$ this one came out with $U = I$ and
$$
S =
\left(
\begin{array}{cc}
2&0 \\
0&4
\end{array}
\right) ,
$$
$$
V =
\left(
\begin{array}{cc}
3&4 \\
2&3
\end{array}
\right) .
$$
This allows us to give an improved basis for $A^\ast,$ while keeping $A$ because $U^{-1} = I$ and $U^{-1}A=A.$ Let vectors $x_1, x_2$ be the rows of
$$
VA^\ast =
\left(
\begin{array}{ccccc}
\frac{1}{2}&\frac{1}{2}&0&0&1 \\
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&\frac{3}{4}
\end{array}
\right)
$$
as in the proof of Lemma (2.3) in\cite{looijenga},
while  vectors $a_1, a_2$ are the rows of
$$
U^{-1}A =
\left(
\begin{array}{ccccc}
1&1&0&0&2 \\
1&1&1&0&3
\end{array}
\right) .
$$
Note that the basis vectors are aligned. We have $a_1 = 2 x_1$ and $a_2 = 4 x_2.$ The quotient group $M^\ast / M  $ is therefore $Z/2Z \oplus Z/4Z.$ This follows the method on page 36 of Newman~\cite{newman}. We also need vectors $z_1, z_2$ in the ambient lattice such that $z_i \cdot a_j = x_i \cdot a_j. $ This was easy enough, I took $z_1, z_2$ as the rows of
$$
Z =
\left(
\begin{array}{ccccc}
0&1&0&0&1 \\
1&1&1&0&0
\end{array}
\right)
$$
We are finally ready to solve the linear Diophantine system $A B^T = 0,$ where $B$ will be a basis matrix for the full rank lattice $M^\perp.$ 
Alright, our lattice $M$ is primitively embedded. We take the definition to be that the matrix $A$ can be completed to become a square matrix of integers with determinant $1.$ What we want is to follow\cite{gilbert} and use column operations to force $A$ into Hermite Normal Form. We find integer matrix $R$ such that $\det R = 1,$
$$
R =
\left(
\begin{array}{ccccc}
1&0&1&0&-2 \\
0&0&-1&0&0 \\
-1&1&0&0&-1 \\
0&0&0&1&0 \\
0&0&0&0&1 \\
\end{array}
\right).
$$
we get
$$
AR = 
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&1&0&0&0
\end{array}
\right)
$$
So, what is a basis for $M^\perp \; ?$ We know that an integer column vector $X$ that solves $ARX =0$ can be any
$$
X = 
\left(
\begin{array}{c}
0 \\
0 \\
p \\
q \\
r \\
\end{array}
\right)
$$
Thus $RX$ can be any integer linear combination of the three right columns of $R.$ Writing as rows, we find a basis of 
$M^\perp \;$ given by
$$
B = 
\left(
\begin{array}{ccccc}
1&-1&0&0&0 \\
0&0&0&1&0 \\
-2&0&-1&0&1
\end{array}
\right)
$$
The Gram matrix for $B$ is
$$
H = 
\left(
\begin{array}{ccc}
2&0&-2 \\
0&1&0 \\
-2&0&6
\end{array}
\right)
$$
which has determinant $8,$ same as $\det G.$ Who Knew?
