A minor omission from the reduction algorithm concerns the final diagonal matrix, which is not necessarily the identity matrix, but may incorporate (an even number of) -1 entries along the main diagonal. It's unclear to me whether these might somehow be absorbed into rotations earlier in the sequence; otherwise they require an extra floor(n/2) rotations to eliminate.

--of course, they can be incorporated into rotations earlier. There of course are always exactly 2 choices for a Givens 2x2 each time, + or -.

WDS raised the question how many length m = n_C_2 sequences of Givens' rotations R_ij are "valid", reducing a general orthonormal matrix to diagonal. Denoting this number f(n) , for n = 1,...,5 , I compute f(n) = 1, 1, 3, 50, 6821; for example for n = 3 , there is a choice of [[2, 1], [3, 1], [3, 2]] , [[3, 1], [2, 1], [3, 2]] , [[3, 2], [2, 1], [3, 1]] , (restricted always to the lower triangle).

--you're wrong again. You perhaps have solved some other problem however, and maybe someday you should figure out what it is. Rewriting your n=3 examples as orderings 1 2& 3* 2 1 3* 2 3&* 1 we see immediately that not a single one is valid, they each destroy previously created zeros at the * steps, and generically fail to zero that entry at all in the & steps. (And don't bother telling me your flawless computer program has shown the contrary. It is simply wrong. We already did that.)

I slipped up there --- thankyou. Fortunately for me, Maple gets it right. WFL

--wrong again, already pointed out your maple program was obviously wrong.

(1) Steps A0 to A2 operate on rows 3 and 4, not 1 and 2 .

--obviously wrong since each step operates on two rows, not 1 row. You are just massively massively utterly wrong and confused about everything you are saying, over and over again, but say it with such a tone of confidence, and never admit there was a problem. Maybe you should run for US president? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)