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Here is an intriguing fact:

Take any $3\times 3$ orthogonal matrix $A$. For example, take $$ A=\begin{pmatrix} 1/\sqrt{14} & \sqrt{2/7} & 3/\sqrt{14} \\ 2 \sqrt{13/105} & -5 \sqrt{5/273} & 8/\sqrt{1365} \\ \sqrt{13/30} & \sqrt{10/39} & -11/\sqrt{390} \end{pmatrix} $$ which has been produced by applying the Gram–Schmidt process to some random choice of vectors.

Compute the determinant of any $2\times 2$ submatrix. For example, the one consisting of the four corners: $$ \begin{vmatrix} 1/\sqrt{14}& 3/\sqrt{14} \\ \sqrt{13/30} & -11/\sqrt{390} \end{vmatrix} = -5 \sqrt{5/273} $$ Note that the result is the middle element of $A$.

Doing this for any submatrix of any $A$, we always get $\pm$ the entry that completes the submatrix. Why is that?

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Given a square matrix $M \in SO_n,$ so $\det M = 1$ and $MM^T = M^T M = I,$ 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:

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

Let's see; given any submatrix defined by a choice of $p$ rows and $p$ columns, we can make a "dual" submatrix of the other $n-p$ rows and $n-p$ columns. By permuting the row and permuting the columns, we can make the $p$ by $p$ matrix be the upper left square corner and the other matrix the lower right. I suppose some more care is needed about $\pm$ signs; so far this just shows that the determinants of the originals have the same absolute value. EDIT: this is as it should be. If we take an off-diagonal element in the matrix of the question, for instance, the 1,2 position with $\sqrt{2/7},$ the determinant of the submatrix with positions 2,1; 2,3; 3,1; 3,3 is negative. More generally, we can make some row transpositions and some column transpositions, always one of the $p$ switched with one of the $n-p$ so as to keep the submatrices intact, and get the $p$ by $p$ as the upper left corner, having not changed the determinants. However, if the total number of transpositions was odd, we end up with $\det M = -1$ and $\det A = - \det D.$

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  • $\begingroup$ I understand why this single line implies that $\det A=\det D$, but I don't see how it was obtained. $\endgroup$
    – user303201
    Jan 6 '16 at 20:45
  • $\begingroup$ Never mind, I get it now. Thanks for the answer ! $\endgroup$
    – user303201
    Jan 6 '16 at 20:48
  • $\begingroup$ I came up with the fact when going through Warner's differential geometry book in graduate school. I showed it to a faculty member who gave me the short proof the next day. He was an expert on matrices, I'm afraid i did not write down his name anywhere and do not remember. $\endgroup$
    – Will Jagy
    Jan 6 '16 at 20:48
  • $\begingroup$ Its a cool trick. $\endgroup$
    – user303201
    Jan 6 '16 at 20:49
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The inverse of a matrix is given by: $$ (A^{-1})_{ij}=\frac{1}{\det A}A'_{ji} $$ where $A'_{ji}$ is the signed minor of the element $a_{ji}.$ Now, for an orthogonal matrix, we have $$ Q'_{ji}=\det Q (Q^{-1})_{ij}=\pm (Q^T)_{ij}=\pm Q_{ji}. $$

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