Let $A\in M_{n \times k}(\mathbb{R})$. Show that $\det(A^TA)=\sum \det(B^2)$ where the sum runs through all $k \times k$ submatrices $B$ of $A$.

Question

Let $A\in M_{n \times k}(\mathbb{R})$. Show that
\begin{equation}
\det(A^TA)=\sum \det(B^2) \;\;\;\;\;\;\;(*)
\end{equation}
where the summation runs through all $k \times k$ submatrices $B$ of $A$ that result from deleting rows of $A$.

Proof attempt:
If $n<k$, then $A^TA$, a $k \times k$ matrix, has rank at most $n<k$. Thus, it is singular and has determinant 0. Also, since $n<k$, there are no $k \times k$ submatrices of $A$ to consider. Thus, both sides of (*) are 0.

If $n=k$ the only $k \times k$ submatrix of $A$ is itself so the right hand side of (*) becomes $$\det(A^2)=\det(A)\det(A)=\det(A^T)\det(A)$$ which is equal to the left hand side.

However, I cannot seem to figure out the case for when $n>k$. I would appreciate an approach or a reference for a solution.

Answer 1

Spoiler alert:

this is the Cauchy-Binet formula in the special case $B=A^T$.