2C12=-2C12(no new information from row one) C22+ 2C32=-2C22(row two) C22=-2C32(again the third row offers no new information) C122+C222+C322=1(from normalization) C12+(-2C32)2+ C122+5C 12=(1-5C32)(Note: again, two equations in three unknowns C11C12+ C21 C22+ C3 C32=0( from orthogonalization) Now there are five equations with six unknowns Arbitrarily choose C11=0 (whenever there are degenerate eigenvalues, there are not unique eigenvectors because he degenerate eigenvectors span a 2-or more-dimensional space, not two unique directions. One al ways is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors) C11=0=y1-5C C31=V02 C21=-2V0.2 C1l C12+ C21 C22+C31 C32=0(from orthogonalization) 0+-2V0.2(-2C32)+V0.2C32=0 5C32=04 -2C12 = -2C12 (no new information from row one) -C22 + 2C32 = -2C22 (row two) C22 = -2C32 (again the third row offers no new information) C122 + C222 + C322 = 1 (from normalization) C122 + (-2C32) 2 + C322 = 1 C122 + 5C322 = 1 C12 = (1- 5C32 2 ) 1/2 (Note: again, two equations in three unknowns) C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) Now there are five equations with six unknowns. Arbitrarily choose C11 = 0 (whenever there are degenerate eigenvalues, there are not unique eigenvectors because the degenerate eigenvectors span a 2- or more- dimensional space, not two unique directions. One always is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors). C11 = 0 = 1 - 5C312 5C312 = 1 C31 = 0.2 C21 = -2 0.2 C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) 0 + -2 0.2(-2C32) + 0.2 C32 = 0 5C32 = 0