real orthogonal matrices

General comments about real-orthogonal matrices PhL 4.10.12 See Theorem at the end. For any real orthogonal matrix we have: RRT = 1 => det(R) = ± 1 The Case N = 2. In Section 9 of my old matrix notes, I start with M = and note that M-1 = /detM For a real orthogonal matrix we must then have MT = M-1 so = / (ad-bc) or (ad-bc) = or (ad-bc)a = d 1 (ad-bc)c= -b 2 (ad-bc)b = -c 3 (ad-bc)d = a 4 How would you solve this set of 4 equations for 4 unknowns? Divide 2/3 to find that c/b = b/c => c2 = b2. Divide 1/4 to find that a/d = d/a => a2 = d2 Divide 3/4 to find that b/d = -c/a => ab = -cd Divide 2/4 to find that c/d = -b/a => ac = -bd Divide 1/2 to find that a/c = -d/b => ab = -cd but we already know this from 3/4 Divide 1/3 to find that a/b = -d/c => ac = -bd but we already know this from 2/4 So we now have four simpler equations in four unknowns c2 = b2 a2 = d2 ab = -cd ac = -bd Well write c = σ1b σ1 = ± 1 a = σ2d σ2 = ± 1 Then the last two equations say σ2db= - σ1bd => σ2 = -σ1 σ2d σ1b = -bd => σ1σ2= -1, same as above So eliminate σ2 so we have c = σ1b a = -σ1d This is all that is left. You can pick b and d freely. The final form then is M = So here are the two possible solutions M = detM = -(d2+b2) M = detM = d2+b2 Now if we want detM = ±1 we can set d = cosθ and b = sinθ to get M2 = detM = -1 M1 = detM = +1 So these ARE the solutions to this problem. Now easy to show that = Now causes a reflection in the y-axis causing y→y but x→-x. Call this ry. Then M1 = ry M2 and M2 = ry M1 since ry2 = 1 So our second solution is to first reflect in the y axis, THEN to rotate. Note that this is not a parity transformation! So consider: ry = rx = = -ry So consider again: det = 1 solution: M1 = only 1 free parameter det = -1 solution: M2 = rxM1 = = M3 = ryM1 = - rxM1 = - M2 M4 = M1rx = = M5 = M1ry = -M4 Now the M4 case arises from θ→-θ on the M2 case, so if we think of θ in (-π,π), then M4 is then not a new solution. Similarly, if we take M2 and let θ → π-θ, we change the cosine sign only, but then we could do the other thing and change the sin only. So in effect there is only one det = -1 solution. Conclusion: for N = 2, one solution is the usual rotation matrix, while the other solution is that rotation combined with either rx or ry. The Case N = 3. It seems clear that one solution here is the usual rotation matrix, while all the other solutions with det = -1 come by combining this with combinations of rx, ry, rz, or P. Note that rxry = rzP So these are "combinations of reflections in axes". For N = odd, P is included in the set. Theorem: If we are told that N is a real orthogonal matrix, we know right away that detN = ± 1. When det = +1, our matrix N is a rotation matrix possibly combined with an even number of axis reflections. When det = -1, the matrix is a rotation matrix combined with an odd number of axis reflections. When N = odd, this includes the parity transformation where we reflect all axes. The reflections affect the determinant in the obvious manner, but don't affect the fact that MMT = 1. Here is why (based on the fact that ririT =1 since a simple diagonal matrix of ±1's ) M' ≡ r1r2.. M r3r4.... M'T ≡ .....r4Tr3T MT ....r2Tr1T MM'T ≡ r1r2.. M r3r4.... .....r4Tr3T MT ....r2Tr1T = r1r2.. M MT ....r2Tr1T = r1r2.. 1....r2Tr1T = 1