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. 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). If complete columns are eliminated in left-to-right order, there is an independent choice of up or down row order available for each, leading to (weak) lower bound f(n) >= 2^(n-2) . Apparently for each k , the final k_C_2 moves R_ij always have i >= n+2-k ; assuming this yields an evan more feeble upper bound, maybe f(n) <= (m!)/2 ? For n = 4 , the diagonal sequence with consecutive [i, j] = ..., [2, 1], [3, 2], [4, 3], ... never occurs; from this I think it follows that no diagonal triad occurs for any n . Fred Lunnon