A curious property of orthogonal matrices

Let $A$ be an $n\times n$ orthogonal matrix, i.e. $AA^T=A^TA=I$. I noticed experimentally that if we take any increasing set of indices $i_1<\cdots<i_p$, then the sum of the squares of all possible minors with rows corresponding to $\{i_1,\ldots,i_p\}$ is equal to 1.

This is a bit hard to formulate, so let me try to say what I mean more explicitly. Let $$A=\begin{pmatrix}a_1^1&\ldots&a_1^n\\\vdots&\ddots&\vdots\\a_n^1&\cdots&a_n^n\end{pmatrix}$$ and let $$A_{i_1\ldots i_p}^{j_1\ldots j_p}=\begin{pmatrix}a_{i_1}^{j_1}&\ldots&a_{i_1}^{j_p}\\\vdots&\ddots&\vdots\\a_{i_p}^{j_1}&\cdots&a_{i_p}^{j_p}\end{pmatrix}$$ for $i_1<\cdots<i_p$, $j_1<\cdots<j_p$ and $1\leq p\leq n$.

Conjecture: If $A$ is orthogonal, then, for any $i_1<\cdots<i_p$ we have $$\sum_{j_1<\cdots<j_p}\left(\det A_{i_1\ldots i_p}^{j_1\ldots j_p}\right)^2=1,$$ where the sum is over all increasing sets of $p$ indices.

How do we prove such a thing? Is that even true? (The answer is very likely yes, as I have checked it for many random examples.)

Note: The case $p=1$ is the statement that the rows of $A$ have unit norms, and the case $p=n$ is the statement that $(\det A)^2=1$. The conjecture generalizes these two statements.

In particular, the conjecture is clearly true for $n=2$. Now, here is a slightly less trivial example:

Example: Take $$ A= \begin{pmatrix} 1/\sqrt{2}&0&1/\sqrt{2}\\ -1/\sqrt{6}&\sqrt{2/3}&1/\sqrt{6}\\ 1/\sqrt{3}&1/\sqrt{3}&-1/\sqrt{3} \end{pmatrix}\in O(3). $$ Then, with $p=2$ and $i_1=1$, $i_2=2$ we have $$ \begin{vmatrix} 1/\sqrt{2}&0\\ -1/\sqrt{6}&\sqrt{2/3} \end{vmatrix}^2 + \begin{vmatrix} 1/\sqrt{2}&1/\sqrt{2}\\ -1/\sqrt{6}&1/\sqrt{6} \end{vmatrix}^2 + \begin{vmatrix} 0&1/\sqrt{2}\\ \sqrt{2/3}&1/\sqrt{6} \end{vmatrix}^2 =\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1. $$

Yes, your conjecture is true. Given the inner product on $\Bbb R^n$, there are induced inner products on $\bigwedge^p\Bbb R^n$. Since the vectors $v_1,\dots,v_p$ formed by your $p$ rows are mutually orthogonal unit vectors, with respect to that inner product, $\|v_1\wedge\dots\wedge v_p\|^2=1$, as well, and that's what you're computing. Perhaps a good way to prove your conjecture is to prove that the quantity you are considering is invariant under orthonormal change of basis.

Following the Darij's comment. Let $B$ be the $p\times n$ submatrix of $A$ consisting of the lines $i_1,\cdots,i_p$ of $A$. Then $BB^T=I_p$ and $\det(BB^T)=1$. Thus $V=\sqrt{1}=1$ is the $p$-dimensional volume of the parallelotope $\Pi$ spanned in $\mathbb{R}^n$ by the $p$ rows of $B$. According to Cauchy Binet, $V^2$ is also the sum of the squares of the volumes of the orthogonal projections of $\Pi$ onto the $\binom{n}{p}$ $p$-dimensional coordinate planes.