Equality of the determinants of certain submatrices of an orthogonal matrix Is the determinant of any submatrix of an ORTHOGONAL matrix extracted from the intersection of $k$ row and $k$ columns equal to that of the $(n-k)(n-k)$ submatrix remaining after deletion of these rows and columns (up to a sign)?! this is something I arrived at in trials to show that the Hodge Star operator’s definition does not depend on the choice of an orthonormal basis of $T*_xM$, $M$ a Reimannian manifold.
 A: Given a square matrix $M \in SO_n$ decomposed as illustrated  with square blocks $A,D$ and rectangular blocks $B,C,$
$$M  = \left( \begin{array}{cc} 
A & B \\\  
 C & D 
 \end{array} \right) ,$$
then   $\det A = \det D.$
What this says is that, in Riemannian geometry with an orientable manifold, the Hodge star operator is an isometry, a fact that has relevance for Poincare duality.
http://en.wikipedia.org/wiki/Hodge_duality
http://en.wikipedia.org/wiki/Poincar%C3%A9_duality
But the proof is a single line:
$$ \left( \begin{array}{cc}  A & B \\\  0 & I  \end{array} \right) 
\left( \begin{array}{cc}  A^t & C^t \\\ B^t & D^t  \end{array} \right)   = 
\left( \begin{array}{cc}  I & 0 \\\  B^t & D^t  \end{array} \right). $$
A: Let $M\in{\bf GL}_n(k)$ be written blockwise
$$M=\begin{pmatrix} A & B \\\\ C & D \end{pmatrix},$$
with $A$ a square, invertible, submatrix. The Sherman-Morrison formula says that
$$\det M=\det A\cdot\det(D-CA^{-1}B).$$
On the other hand, it can be shown that
$$M^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (D-CA^{-1}B)^{-1} \end{pmatrix}.$$
We therefore have
$$\det M\cdot\det(D-CA^{-1}B)^{-1}=\det A.$$
When $M$ is orthogonal, one has $\det M=\pm1$, and the previous formula is exactly what you guessed.
