I am aware that an arbitrary $3D$ Rotational Matrix can be constructed by successive rotations about the $x, y$ and $z$ axes. How can I use this to show that all $3D$ rotation matrices are othogonal and have $det=1$.
Would I just need to individually prove each matrix is orthogonal, i.e show that the transpose of each matrix is the same as its inverse and then show that the product of the matrices is also orthogonal?
For showing that the $det=1$ is this just a trivial calculation of the determinant of each rotation matrix? And then using the property that the determinant of the product of matrices is just the product of each individual determinant?
Best Answer
You obtain your rotation matrix $R$ from the product of three matrices
$$R=R_x(\alpha)\,R_y(\beta)\,R_z(\gamma).$$
If $R $is orthogonal thus $R^T\,R=I_3$ where $I_3$ $3\times 3$ identity matrix
$$R^T\,R=(R_x\,R_y\,R_z)^T\,(R_x\,R_y\,R_z)=R_z^T\,R_y^T\,R_x^T\,R_x\,R_y\,R_z$$
with:
$R_x^T\,R_x=I_3\quad,R_y^T\,R_y=I_3$ and $\quad R_z^T\,R_z=I_3.$
Thus $$R^T\,R=I_3\, ,\quad R^{-1}=R^T\, ,$$
and the determinant of $R$ is:
$$\det(R)=\det(R_x)\,\det(R_y)\,\det(R_z)=1\times 1\times 1=1$$
$$S_x(\alpha)=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0& \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right] $$
$$S_y(\beta)= \left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left( \beta \right) \\ 0&1&0\\ -\sin \left( \beta \right) &0&\cos \left( \beta \right) \end {array} \right] $$
$$S_z(\gamma)= \left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left( \gamma \right) &0\\ \sin \left( \gamma \right) &\cos \left( \gamma \right) &0\\ 0&0&1\end {array} \right] $$