Despite generous previous assistance from all, I am still puzzling over the mysterious behaviour of factorisation into Givens' rotations. My latest head-scratchers are embedded below at (6) -- (8) within in a potted summary of developments to date. I cling to the expectation that these may also be trivial when viewed from the correct perspective ... WFL ______________________________________________________________________ (1): R_pq(u) denotes "Givens'" rotation through angle u in the (x_p, x_q)-coordinate plane about the origin of Euclidean n-space, represented by an order-n orthonormal matrix taking an elementary form: see https://en.wikipedia.org/wiki/Givens_rotation (2): Given(!) natural n , a "schedule" is a permutation of distinct [p, q] with 1 <= q < p <= n , specifying some reduction of a general order-n orthonormal matrix to diagonal form via that sequence of "moves", ie. premultiplications by R_pq(u_k) for k in [1..m] . Note that the matrix remains orthonormal throughout. (3): During reduction of a specific numerical matrix, angles u_k assume individual numerical values, assigned to clear matrix entry [p, q]_k . At each stage k there are two suitable assignments: if u is one, the other is u + pi . For the present angles are unimportant and can be ignored. Sample reduction from orthonormal to diagonal, showing values p, q, A at each stage k for random numerical matrix A : https://www.dropbox.com/s/uwsmq2habm4tfgp/given_p1.tiff (4): Given a schedule along with natural r in [1..n] , its r-th "sub-schedule" is the subsequence excluding moves for which p = r or q = r , and with indices p, q > r decremented by unity. (5): Examples --- Of 6 permutation sequences for n = 3 , just 3 constitute valid schedules: [ [2, 1], [3, 1], [3, 2] ] , [ [3, 1], [2, 1], [3, 2] ] , [ [3, 2], [2, 1], [3, 1] ] . For n = 4 , one schedule and its sub-schedule with r = 3 : [ [4, 3], [3, 1], [2, 1], [4, 1], [3, 2], [4, 2] ] -> [ [2, 1], [4, 1], [4, 2] ] -> [ [2, 1], [3, 1], [3, 2] ] , first in the list above. (6): Claim --- Every sub-schedule of a schedule is a schedule. (7): Claim --- Every schedule is a sub-schedule of some larger. (8): Puzzle --- *** Are these claims obvious? *** (9): Problem --- *** Are these claims true? *** There are numerous implications: in particular, that for each schedule there exists an angle selection strategy guaranteed to reduce every special orthonormal matrix (orientation-preserving isometry) to the identity. (10): A matrix entry [i, j] is "cleared" after move k when partial reduction by the first k moves has guaranteed that it is zero (and will remain so subsequently). (11): Remark --- [p, q]_k is a move of a valid schedule iff after move k-1 , matrix entries [p, j], [q,j] are either both cleared or both uncleared for all j in [1..n] . (12): Remark --- Move [p, q] affects only matrix rows p, q . Suppose clearing entry [r, j] causes columns j, j' to retain a single pair of uncleared entries: say [r', j], [r', j'] , where {r, r'} = {p, q} . Via orthogonality entry [r', j'] in the upper triangle j' > r' also becomes cleared. Iterate ad nauseam? For example, if move [2, 1] reduces column 1 to [1, 0, ..., 0]' , then any uncleared entries on the rest of row 1 must also be cleared in consequence. (13): Via (11) and (12) distinct schedules of order n can in principle be enumerated combinatorially, and their cardinal f(n) calculated; currently known are values n 0 1 2 3 4 5 6 f 1 1 1 3 50 6525 8515479 f/h - 1.00 1.00 0.89 0.78 0.68 0.59 Plainly f(n) <= m! where m = binom(n, 2) == n(n-1)/2 equals the number of distinct moves [p, q] . From the data it seems faintly plausible that asymptotically f(n) ~ c h(n) , where h(n) == (n/2)^m , with constant c < 1/2 by some substantial margin. Fred Lunnon 01/06/16 ______________________________________________________________________