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We can see that in order to minimize J, we need to maximize g(R). This problem can be written in a better and more elegant way by representing rotation matrix R in terms of the quaternion
[[File:Equation29.png|frame|center]]
R can be written in terms of q as<math>R = (q_4^2-q^Tq)I+2qq^T-2q_4[[File:Equation30.png|frame|center]q \times]</math><br \>
Where [q x] is the cross product matrix associated with q
[[File:Equation31.png|frame|center]]
<math>g(\bar{q}) = \bar{q}^TKq</math><br \>
Where K is a 4x4 Matrix given by
[[File:Equation33Equation77.png|frame|center]]Wherein,<br \><math>B = \sum w_k(v_{kb}v_{ki}^T)</math><br \><math>S = B+B^T</math><br \><math>Z = \sum w_k(v_{kb} \times v_{ki})</math><br \><math>\sigma = tr(B)</math> <br \>[[File:Equation34.png|frame|center]]Now we want to maximise g. However, noting that quaternion elements are not independent, the constraint [[File:Equation35.png]] <math>\bar{q}^Tq = 1</math> must also be satisfied. So we use the lagrangian multiplier method and differentiate with respect to [[File:Equation36.png]] <math>\bar{q}</math> we get [[File:Equation37<math>K\bar{q} = \lambda \bar{q}</math>.png]] So now, we have reduced the problem to an eigenvalue problem for the matrix K. Since, K is a 4x4 matrix, it can have at most 4 different eigenvalues. Substituting [[File:Equation37.png]] <math>K\bar{q} = \lambda \bar{q}</math> in [[File:Equation38.png]]<math>g(\bar{q}) = \bar{q}^TKq</math>, we get
[[File:Equation39.png|frame|center]]
Now we want to maximise g. So we will chose the eigenvector corresponding to the maximum eigenvalue.