Sum of squares of $k\times k$ cofactors is $1$ for an orthonormal matrix [closed]

Question

Let $n,k\in \mathbb N$ with $k\leq n$. Let $A$ be an $n\times n$ real orthonormal matrix. Fix any $k$ rows of $A$ and from there consider every possible $k\times k$ cofactors and there will be exactly $n\choose k$ cofactors. Then prove that sum of the squares of those $n\choose k$ cofactors will be equal to $1$.

I considered the case when $n=3$ and $k=2$. For that case, if we write $A=(a_{ij})_{1\leq i,j \leq 3}$, then fixing first and second rows, the sum of the squares of all $2\times 2$ cofactors will be

\begin{align} &(a_{11}a_{22}-a_{12}a_{21})^2+(a_{11}a_{23}-a_{21}a_{13})^2+(a_{12}a_{23}-a_{13}a_{22})^2\\ &=(a_{11}a_{22})^2-2 a_{11}a_{22} a_{12}a_{21}+(a_{12}a_{21})^2+ (a_{11}a_{23})^2-2a_{11}a_{23}a_{21}a_{13}+(a_{21}a_{13})^2\\ & \ \ \ +(a_{12}a_{23})^2-2 a_{12}a_{23}a_{13}a_{22}+(a_{13}a_{22})^2\\ &=1 \end{align} The last equality comes from using the relations that $\displaystyle\sum_{i=1}^{3}{a_{1i}}^2=1$, $\displaystyle\sum_{i=1}^{3}{a_{2i}}^2=1$ and $a_{11}a_{21}+a_{12}a_{22}+a_{13}a_{23}=0$.

I just want help to generalize this result for any $k\times k$ cofactors. How to do it. Any help is appreciated. Thank you.

Answer 1

I just want to add a proof that may be given independently: Suppose we are given $k$ mutually orthogonal vectors in ${\mathbb{R}}^n\backslash\{0\}$; in matrix form this is a $k\times n$ matrix $A=[a_{i,j}]$. A $k$ choice of $k$ columns indices from $n$ is denoted by $\sigma_i$.

The minor $M_{k,\sigma_i}$ is the minor formed by the columns with indices in $\sigma_i$ and rows $1$ to $k$.

The minor $M_{k-1,\sigma_i(\tilde{j})}$ is the minor formed by the columns with indices in $\sigma_i$ except $j$ ($j\in\sigma_i$) and rows $2$ to $k$.

For $n\ge 2$ fixed we want to prove $$(1) \quad \sum_{i=1}^{\binom{n}{k}}(M_{k,\sigma_i})^2=\prod_{i=1}^k(\sum_{j=1}^n a_{i,j}^2)$$ The cases $k=1$ and $k=2$ are either easy or trivial, suppose this is true for $k=r$, we want to prove it for $k=r+1$.

Expand the left hand side of $(1)$ with respect to the first row $$\begin{aligned}(2)\quad \sum_{i=1}^{\binom{n}{k}}(M_{k,\sigma_i})^2&=\sum_{i=1}^{\binom{n}{k}} \big(\sum_{j\in\sigma_i}a_{[1,j]}M_{k-1,\sigma_i(\tilde{j})} \big)^2\\&=\sum_{i=1}^{\binom{n}{k}}\big(\sum_{j\in\sigma_i}(a_{[1,j]}M_{k-1,\sigma_i(\tilde{j})})^2+2\sum_{\,\,j\neq j'\\j\&j'\in\sigma_i}a_{[1,j]}a_{[1,j']}M_{k-1,\sigma_i(\tilde{j})}M_{k-1,\sigma_i(\tilde{j'})}\big)\end{aligned}$$ where $a_{[1,j]}$ is the entry $a_{1,j}$ with the corresponding sign in $M_{k,\sigma_i}$

Set $M_{k-1,s_i}$ as the minor formed by rows $2$ to $k$ and columns indices in $s_i$ ($|s_i|=k-1$). The idea is to make the remaining $M_{k-1,s_i}$ appear in $(2)$ so that we get $$(3)\quad(\sum_{j=1}^n a_{1,j}^2)(\sum_{i=1}^{\binom{n}{k-1}}M_{k-1,s_i}) $$ and from the induction hypothesis we can conclude.

Take the system on $k-1$ equations $$(4)\quad\sum_{j=1}^n a_{1,j}a_{h,j}=0\quad h=2,\ldots k$$

A choice of $k-2$ column indices is denoted by $t_i$. The set $t_i(l) $ ($l\notin t_i$) is the set $t_i$ with index $l$ added so $t_j(l)=s_i $ for some $i,j$. Multiply each equation in the system by a $k-2$ minor $r_h$ from columns indices in $t_i$ so that $$(5) \quad\sum_{h=2}^{k}r_h\sum_{j=1}^n a_{1,j}a_{h,j}=\sum_{l,l\notin t_i}\epsilon_l a_{1,l}M_{k-1,t_i(l)}=0$$ where, $\epsilon_l=\pm 1$. Squaring the last equation for every choice $t_i$ and summing up we see that the number of product terms (with factor $2$) is $\binom{n}{k-2}\times \binom{n-k+2}{2}$ and this is equal to the number of product terms in $(2)$ or $\binom{n}{k}\times \binom{k}{2}$.

Finally notice that for every $t_i$, $l\neq l'$ and $l\&l'\notin t_i$ we have $$ \epsilon_{l'}\epsilon_l a_{1,l}a_{1,l'}M_{k-1,t_i(l)}M_{k-1,t_i(l')}=-a_{[1,l]}a_{[1,l']}M_{k-1,\sigma_{i_{l,l'}}(\tilde{l})}M_{k-1,\sigma_{i_{l,l'}}(\tilde{l'})}$$ with $t_i(l)=\sigma_{i_{l,l'}}(\tilde{l'}),t_i(l')=\sigma_{i_{l,l'}}(\tilde{l})$.

Replacing the product terms by the square ones (after squaring $(5)$ and summing over $t_i$) in $(2)$ we obtain $(3)$.