Sum of squares of maximal minors of a rectangular matrix with orthonormal rows A matrix $A$ has $m$ rows and $n$ columns, such that $m \leq n$. We know that each row of $A$ has norm $1$ (the norm of an element $x=(x_1,x_2,...,x_n) \in \mathbb{R}^n$ is $||x||=\sqrt{x_1^2+x_2^2+...+x_n^2}$) and any two rows are  orthogonal (if $y=(y_1,y_2,...,y_n) \in \mathbb{R}^n$ then $x$ and $y$ are orthogonal if their dot product is $0$, i.e. $x_1y_1+x_2y_2+\cdots+x_ny_n=0$. Is the sum of the squares of the minors of order $m$ of $A$ equal $1$? I have already asked this question before, but in a cheeky way, so nobody gave it a try. I think that this is true.
What I did, was to find out that $A^tA=I_n$ and I know that the eigenvalues in case $m=n$ have absolute value 1 (they are not necessarily real). Observation: The matrix is not a square matrix. We have $A^tA=I_n$ but $^tAA=I_n$ is not equivalent to the first (if this relation holds, then $n\leq m$ which is not given).
Any ideas?
 A: This is indeed true, as Care Bear has shown using the Cauchy-Binet formula. One should also be able to show this directly (with significantly more work!) by multiplying the $m$ equations $(x_{11}^2 + \ldots + x_{1n}^2) = \ldots = (x_{m1}^2 + \ldots + x_{mn}^2) = 1$ together, expanding to a homogeneous polynomial of degree $2m$, and using the orthogonality relations. For yet another way, here is a more theoretical proof:
Viewing $A$ as a linear map $A : \mathbb{R}^n \to \mathbb{R}^m$, $n \ge m$, the size $m$ minors of $A$ are precisely the entries of $\wedge^m A : \wedge^m \mathbb{R}^n \to \wedge^m \mathbb{R}^m \cong \mathbb{R}$, represented by a $1 \times {n \choose m}$ matrix. Now $A^T$ is the matrix of the dual map $A^T : (\mathbb{R}^m)^* \to (\mathbb{R}^n)^* \cong \mathbb{R}^n$. Since the rows of $A$ are orthonormal, $AA^T = I_m$, so since $\wedge$ is functorial and commutes with duals, $(\wedge^m A)(\wedge^m A)^T = (\wedge^m A)(\wedge^m A^T) = \wedge^m(AA^T) = \wedge^m (I_m) = I_1$, which says exactly that $\wedge^m A$, viewed as a row vector, has norm $1$.
A: 
the eigenvalues of $A$ have absolute value $1$ 

To my knowledge, the concept of eigenvalues applies to square matrices only. 

What I did, was to find out that $A∗^tA=I_n$

This can't be true if $m<n$, since the rank of $A$ is less than $n$. What is true is that $AA^t=I_m$; this says precisely that the rows of $A$ are orthonormal. 
Your conjecture about the sum of squares of $m$-minors is correct: they add up to $1$. Indeed, this sum is equal to $\det (AA^t)$ by the Cauchy–Binet formula. 
