# 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.

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Yes, and you arrived at this exactly the way I did. I mentioned it to other differential geometry people, it was unfamiliar to all. But I also showed it to a matrix expert in the late 1980's, he showed me a one-line proof, typeset in my answer below. Note that column permutations and row permutations allow you to take a standard block pattern without loss of generality. – Will Jagy Jun 23 '13 at 17:32

## 2 Answers

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:

$$\left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right).$$

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I first put this at mathoverflow.net/questions/3134/… in March 2010. So sue me. – Will Jagy Jun 23 '13 at 17:26

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.

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Recent work (2012) of Richard Brent and Judy-ann Osborn on determinants of binary matrices uses a similar (if not identical) observation by F. Szollozi to get better bounds on such determinants. Gerhard "Ask Me About Binary Matrices" Paseman, 2013.06.23 – Gerhard Paseman Jun 23 '13 at 16:56
Denis, in this special case there is a one line proof. I cannot recall the name of the guy who found it for me, he was a matrix (and other) expert at U.C. San Diego in the late 1980's. – Will Jagy Jun 23 '13 at 17:35