Let A be a matrix in SO3(R). The characteristic polynomial of A is of degree three, and every real polynomial of odd degree has a real root. Therefore, A has a real eigenvalue, λ. Let v be a normalized eigenvector corresponding to λ. Then |λ| = ||Av|| = ||v|| = 1, so λ = ± 1. If S denotes the orthogonal complement of v, then for all x in S, <Ax, v> = <Ax, ± Av> = ± <x, v> = 0. This shows that S is an A-stable subspace. Hence, there is an invertible matrix P such that P-1AP is in block diagonal form [λ] ⊕ A' where A' is in O2(R). If λ = 1, we are done. If λ = -1, then det A' = -1, forcing the eigenvalues of A to be -1 or 1. In either case, A has eigenvalue 1.
The result does not hold for SO2(R), e.g., the matrix of plane rotation about the origin by 90º counterclockwise (i.e., the matrix A = {{0, -1}, {1,0}}) is an element of SO2(R) that does not even have real eigenvalues.
