A matrix $A$ has $m$ rows and $n$ colums, such that $m \leq n$. We know that each row of $A$ has the 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. $x_1y_1+x_2y_2+...+x_ny_n=0$. Is the sum of the squares of the minors of order $m$ of $A$ equal $1$? I think that this is true. What I did, was to find out that $A*^tA=I_n$ and I know that the eigenvalues of A have absolute value 1 (they are not necessarily real). Observation: The matrix is not a square matrix. We have $A*^tA=I_n$ but $^tA*A=I_n$ is not equivalent to the first (if this relation holds, then $n\leq m$ which is not given).