Euler's theorem (rotation)

In mathematics, Euler's theorem for rotations states that a rotation of a three-dimensional rigid body (a motion of the rigid body that leaves at least one point of the body in place) is around an axis, the rotation axis. This means that all points of the body that lie on the axis are invariant under rotation, i.e., do not move.

Proof

Leonhard Euler gave a geometric proof that rests on the fact that a rotation can be described as two consecutive reflections in two intersecting mirror planes. Points in a mirror plane are invariant under reflection and hence the points on the intersection (a line) of the two planes are invariant.

An algebraic proof starts from the fact that a rotation is a linear map in one-to-one correspondence with an orthogonal matrix R, i.e, a matrix for which

$\mathbf {R} ^{\mathrm {T} }\mathbf {R} =\mathbf {R} \mathbf {R} ^{\mathrm {T} }=\mathbf {E} ,$

where E is the 3×3 identity matrix and superscript T indicates the transposed matrix. Clearly an orthogonal matrix has determinant ±1, for invoking some properties of determinants, one can prove

$1=\det(\mathbf {E} )=\det(\mathbf {R} ^{\mathrm {T} }\mathbf {R} )=\det(\mathbf {R} ^{\mathrm {T} })\det(\mathbf {R} )=\det(\mathbf {R} )^{2}\quad \Longrightarrow \quad \det(\mathbf {R} )=\pm 1.$

The matrix with positive determinant is a proper rotation and with a negative determinant an improper rotation (is equal to a reflection times a proper rotation).

It is next observed that a rotation matrix R has at least one invariant vector n, i.e., R n = n. If R has more than one invariant vector then R = E and any vector is an invariant vector. Note that this statement is equivalent to stating that the vector n is an eigenvector of the matrix R with eigenvalue λ = 1.

It is easy to show that the proper rotation matrix R has unit eigenvalue. Using

$\det(-\mathbf {R} )=(-1)^{3}\det(\mathbf {R} )=-\det(\mathbf {R} )\quad {\hbox{and}}\quad \det(\mathbf {R} ^{-1})=1,$

we find

${\begin{aligned}\det(\mathbf {R} -\mathbf {E} )=&\det {\big (}(\mathbf {R} -\mathbf {E} )^{\mathrm {T} }{\big )}=\det {\big (}(\mathbf {R} ^{\mathrm {T} }-\mathbf {E} ){\big )}=\det {\big (}(\mathbf {R} ^{-1}-\mathbf {E} ){\big )}=\det {\big (}-\mathbf {R} ^{-1}(\mathbf {R} -\mathbf {E} ){\big )}\\=&-\det(\mathbf {R} ^{-1})\;\det(\mathbf {R} -\mathbf {E} )=-\det(\mathbf {R} -\mathbf {E} )\quad \Longrightarrow \quad \det(\mathbf {R} -\mathbf {E} )=0\end{aligned}}$

From this follows that λ=1 is a solution of the secular equation, that is,

$\det(\mathbf {R} -\lambda \mathbf {E} )=0\quad {\hbox{for}}\quad \lambda =1.$

From linear algebra we know the general result that an m×m matrix A has m orthogonal eigenvectors if and only if it is normal, that is, if A^TA = AA^T. Clearly, a rotation matrix being normal, the solution λ = 1 belongs to an eigenvector that is a member of an orthogonal set of order three, call the vector n, with R n = n. The line μn for real μ is invariant under R, i.e, μn is a rotation axis. This proves Euler's theorem.