mt1 = {{Cos[ξ] Cos[ϕ] - Sin[ξ] Sin[ϕ] Sin[ψ], Cos[ψ] Sin[ϕ], Cos[ϕ] Sin[ξ] + Cos[ξ] Sin[ϕ] Sin[ψ]},
{-Cos[ξ] Sin[ϕ] - Cos[ϕ] Sin[ξ] Sin[ψ], Cos[ϕ] Cos[ψ], -Sin[ξ] Sin[ϕ] + Cos[ξ] Cos[ϕ] Sin[ψ]},
{-Cos[ψ] Sin[ξ], -Sin[ψ], Cos[ξ] Cos[ψ]}};
mt2 = {{a, b, c}, {b, d, e}, {c, e, h}};
$$mt1 = \pmatrix{a& b& c\\b& d& e\\c& e&h}$$
HereFor mt1$mt1^T.mt2.mt1$ = d
, where mt1 is an orthogonal matrix and mt2
, mt2 is a real symmetric matrix. In principle, the output of Transpose[mt1].mt2.mt1
should be and d is a diagonal matrix, but Mathematica cannot do with eigenvalues of mt2 on the diagonal. How to find mt1 such that it is orthogonal and contains three variables using Mathematica. Any suggestion is appreciated!