Let T be a Linear operator on $\mathbb{R^6}$ given by $T(x_1,x_2,x_3,x_4,x_5,x_6)=(-x_6,x_4,x_1,x_3,x_5,x_2)$
I want to find the determinant of the corresponding matrix... When taking the standard basis for both domain and codomain I am getting the determinant as -1
And if I Fixing the domain basis as standard basis $\{e_1,e_2,...,e_6\}$ and for the codomain the basis is $\{T(e_1),...,T(e_6)\}$ It is clear that this is a basis for the codomain and now I am getting the matrix of the transformation as Identity Matrix whose determinant is 1... Why is that?
You cannot just change the basis of your codomain and expect the determinant to stay the same. This would correspond to $\det(BT)$, where $B$ is the matrix of your basis transformation. Since $\det(BT) = \det(B)\det(T)$, this will change your determinant by the determinant of the basis transformation. What you can do, however, is to change it on both domain and codomain to the same basis. In that case you get $\det(BTB^{-1}) = \det(B)\det(T)\det(B^{-1}) = \det(T)$ and everything works out.