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While experimenting with orthogonal vectors I've noticed the following transformation: If $$ A = \begin{bmatrix}z & r \cr c & B\end{bmatrix} $$ is orthogonal, $z$ and $B$ square, and $I+z$ invertible, then the smaller matrix $$ A_1 = B-c(I+z)^{-1}r $$ is also orthogonal, it is not hard to show. I would like to know:

  1. Reference request: Is this a known transformation? It must be, but I don't see it in books so far.

  2. For the case where $z$ is $1\times1$, it is not hard to show that $\frac{1}{1+z}cr$ is bounded as $z$ tends to $-1$. But in fact, setting $A_1 = B$ in the case where $z=-1$, several numerical examples suggest that the mapping $O(n)\to O(n-1)$, $A\to A_1$, might even be continuous. It preserves inverses, but is not a homomorphism. The case $O(2)\to O(1)$ is constant on the two components and is continuous. So: is the mapping $O(n)\to O(n-1)$, $A\to A_1$, continuous?

  3. Entirely speculative: if it is continuous, is $O(n)\to O(n-1)$ a bundle? At least the inverse image of each point seems to be $S^{n-1}$: Given $A_1\in O(n-1)$ and any unit $1\times n$ row vector $[z\ r]$ the matrix $$ A = \begin{bmatrix} z & r \cr -A_1r^T & A_1(I-\frac{1}{1+z}r^Tr) \end{bmatrix} \quad {\rm or,\ if\ } z=-1, \ \ A = \begin{bmatrix} -1 & 0 \cr 0 & A_1 \end{bmatrix}, $$ is in $O(n)$ and maps to $A_1$.

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  • $\begingroup$ Warning: this may be unrelated. I was excited by some recent (2012) results by Brent, Osborn, et. al. on arXiv regarding determinants of binary matrices. They used a nice result of someone whose name is like Szollozi on unitary matrices. Your question suggested this last result to me, which I think goes back to Jacobi. If I find the link before someone else does, I will post it. Gerhard "Ask Me Not About Smartphones" Paseman, 2012.11.14 $\endgroup$ – Gerhard Paseman Nov 15 '12 at 0:15
  • $\begingroup$ Certainly $O(3)$ is not a bundle over $O(2)$, just for $\pi_1$ reasons. $\endgroup$ – Allen Knutson Nov 15 '12 at 0:26
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    $\begingroup$ Just a note: your $\det A = \pm 1,$ and $\det A \det B = \det z.$ See mathoverflow.net/questions/3134/… $\endgroup$ – Will Jagy Nov 15 '12 at 2:38
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    $\begingroup$ reminds me of Schur complement, somehow.en.wikipedia.org/wiki/Schur_complement $\endgroup$ – Dima Pasechnik Nov 15 '12 at 5:24
  • $\begingroup$ @DimaPasechnik: if you define the space $P(n)=I+O(n)$, this means that $P(n)$ is closed by Schur complementation (exactly like positive semidefinite matrices, or M-matrices --- two other classes of matrices for which all the eigenvalues lie in the right half-plane). This seems an interesting statement; it is new to me. $\endgroup$ – Federico Poloni Nov 15 '12 at 13:07

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