Determinant of the product of two rectangular matrices Let $u_1,\dots,u_n\in \mathbb{R}^n$ and $v_1, \dots, v_n\in \mathbb{R}^n$ be two linearly independent sets of vectors. Suppose $U, V\in \mathbb{R}^{n\times (n-1)}$ are matrices such that $U = [u_2 \cdots u_n]$ and $V = [v_2 \cdots v_n]$ (columns consisting of the vectors except $u_1, v_1$). Then, what is $\det(U^\top V)$?
I know that when the two sets are orthonormal, then $|\det(U^\top V)| = |\langle u_1, v_1 \rangle|$. I also don't know how to prove this special case. The Cauchy-Binet formula doesn't seem to give me anything (or perhaps I don't know how to analyze it).
Edit: I understand that there is probably not a closed form formula when the vectors are arbitrary. I'd still like to know how to prove the special case when the vectors are orthonormal.
 A: Since the vectors can be fully arbitary, there won't be a short formula for this, but you might find this formula come in handy to connect the determinants of $(u_1\;U)$ and $(v_1\;V)$ with that of $U^TV$ if it is invertible:
Lemma: Let $a\in\mathbb{R}$ be a scalar, $b,c\in\mathbb{R}^n$ be vectors and $D\in\mathbb{R}^{n\times n}$ be an invertible matrix, then:
$$\det\begin{pmatrix}
a & b^T \\
c & D
\end{pmatrix}
=\det(D)\left(a-b^TD^{-1}c\right)$$
Proof: Let $D_i$ be the $n\times(n-1)$-matrix obtained by $D$ by deleting the $i$-th column. The $n\times n$-matrix $(c\; D_i)$ obtained by adding the vector $c$ as the first column can be transfered into the matrix $D_i(c)$, where the $i$-th column of $D$ has been replaced by $c$ by swapping colums $i$ times, therefore $\det(c\; D_i)=(-1)^i\det D_i(c)$. By Cramer's rule, we further have $\det D_i(c)=\det(D)(D^{-1}c)_i$ and using Laplace expansion for the first row we obtain:
\begin{align*}
\det\begin{pmatrix}
a & b^T \\
c & D
\end{pmatrix}
&=a\det(D)
+\sum_{i=1}^n(-1)^{i+1}b_i\det(c\; D_i)
=a\det(D)
-\sum_{i=1}^nb_i\det D_i(c) \\
&=\det(D)\left(a-\sum_{i=1}^nb_i(D^{-1}c)_i\right)
=\det(D)\left(a-b^TD^{-1}c\right),
\end{align*}
which concludes the proof. $\square$
Analogously we also have:
$$\det\begin{pmatrix}
D & c \\
b^T & a
\end{pmatrix}
=\det(D)\left(a-b^TD^{-1}c\right),$$
which is the version used in (4+1)D Kaluza-Klein theory in physics for example.
