The general 2-D linear mapping over reals can be represented by a 2x2 matrix of reals: [x1 y1]M = [x2 y2] But this mapping can also be represented by 2 complex numbers A, B, which representation also has 4 real parameters: (x+iy) = A*(x+iy)+B^2*(x-iy), or more succinctly, z=Az+B^2z', where z' is the conjugate of z. (We'll see the reason for using B^2 instead of B shortly.) Now the matrix corresponding to (A,B) is symmetric iff A is real, so the eigenvalues of this mapping are real: lambda1,2 = A+-|B|^2 = A+-BB' We could go to the trouble of solving for the eigenvectors, but they are completely obvious (& obviously orthogonal): B, iB. Consider B: A(B)+B^2(B)' = AB+BBB' = AB+(BB')B = (A+BB')B = lambda1 B Consider iB: A(iB)+B^2(iB)' = A(iB)-BBiB' = A(iB)-BB'(iB) = (A-BB')iB = lambda2 B --- I'm sure that something along these lines can be done with quaternions; Cayley and/or Hamilton must have already done this, but it's difficult to decode their writings. More to come.