Euler's theorem (rotation)

In mathematics, Euler's theorem for rotations states that a rotation of a three-dimensional rigid body—which is a motion of the rigid body in 3-dimensional point space 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. In other words, if the motion leaves one point of the rigid body invariant, then it leaves a whole axis through the body invariant.

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 under the two consecutive reflections and hence under rotation.

An algebraic proof starts from the fact that a rotation is a linear map in one-to-one correspondence with a 3×3 rotation 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 a rotation 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 will now be shown that a rotation matrix R has at least one invariant vector n, i.e., R n = n. Note that this is equivalent to stating that the vector n is an eigenvector of the matrix R with eigenvalue λ = 1.

A proper rotation matrix R has at least one unit eigenvalue. Using the two relations:

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

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 root (solution) of the secular equation, that is,

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

In other words, the matrix R − E is singular and has a non-zero kernel, that is, there is at least one non-zero vector, say n, for which

$(\mathbf {R} -\mathbf {E} )\mathbf {n} =\mathbf {0} \quad \Longleftrightarrow \quad \mathbf {R} \mathbf {n} =\mathbf {n}$

The line μn for real μ is invariant under R, i.e, μn is a rotation axis. This proves Euler's theorem.

Equivalence of an orthogonal matrix to a rotation matrix

A proper orthogonal matrix is equivalent to

$\mathbf {R} \sim {\begin{pmatrix}\cos \phi &-\sin \phi &0\\\sin \phi &\cos \phi &0\\0&0&1\\\end{pmatrix}},\qquad 0\leq \phi \leq 2\pi .$

If R has more than one invariant vector then φ = 0 and R = E. Any vector is an invariant vector of E.

Excursion into matrix theory

In order to prove the previous equation some facts from matrix theory must be recalled.

An m×m matrix A has m orthogonal eigenvectors if and only if A is normal, that is, if A^†A = AA^†. This result is equivalent to stating that normal matrices can be brought to diagonal form by a unitary similarity transformation:

$\mathbf {A} \mathbf {U} =\mathbf {U} \;\mathrm {diag} (\alpha _{1},\ldots ,\alpha _{m})\quad \Longleftrightarrow \quad \mathbf {U} ^{\dagger }\mathbf {A} \mathbf {U} =\operatorname {diag} (\alpha _{1},\ldots ,\alpha _{m}),$

and U is unitary, that is,

$\mathbf {U} ^{\dagger }=\mathbf {U} ^{-1}.$

The eigenvalues α₁, ..., α_m are roots of the secular equation. If the matrix A happens to be unitary (and note that unitary matrices are normal), then

$\left(\mathbf {U} ^{\dagger }\mathbf {A} \mathbf {U} \right)^{\dagger }=\mathrm {diag} (\alpha _{1}^{*},\ldots ,\alpha _{m}^{*})=\mathbf {U} ^{\dagger }\mathbf {A} ^{-1}\mathbf {U} =\mathrm {diag} (1/\alpha _{1},\ldots ,1/\alpha _{m})$

and it follows that the eigenvalues of a unitary matrix are on the unit circle in the complex plane:

$\alpha _{k}^{*}=1/\alpha _{k}\quad \Longleftrightarrow \alpha _{k}^{*}\alpha _{k}=|\alpha _{k}|^{2}=1,\qquad k=1,\ldots ,m.$

Also an orthogonal (real unitary) matrix has eigenvalues on the unit circle in the complex plane. Moreover, since its secular equation (an mth order polynomial in λ) has real coefficients, it follows that its roots appear in complex conjugate pairs, that is, if α is a root then so is α^∗.

After recollection of these general facts from matrix theory, we return to the rotation matrix R. It follows from its realness and orthogonality that

$\mathbf {R} \mathbf {U} =\mathbf {U} {\begin{pmatrix}e^{i\phi }&0&0\\0&e^{-i\phi }&0\\0&0&1\\\end{pmatrix}}$

with the third column of the 3×3 matrix U equal to the invariant vector n. Writing u₁ and u₂ for the first two columns of U, this equation gives

$\mathbf {R} \mathbf {u} _{1}=e^{i\phi }\,\mathbf {u} _{1}\quad {\hbox{and}}\quad \mathbf {R} \mathbf {u} _{2}=e^{-i\phi }\,\mathbf {u} _{2}$

If u₁ has eigenvalue 1, then φ= 0 and u₂ has also eigenvalue 1, which implies that in that case R = E.

Finally, the matrix equation is transformed by means of a unitary matrix,

$\mathbf {R} \mathbf {U} {\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {i}{\sqrt {2}}}&0\\{\frac {1}{\sqrt {2}}}&{\frac {-i}{\sqrt {2}}}&0\\0&0&1\\\end{pmatrix}}=\mathbf {U} \underbrace {{\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {i}{\sqrt {2}}}&0\\{\frac {1}{\sqrt {2}}}&{\frac {-i}{\sqrt {2}}}&0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\{\frac {-i}{\sqrt {2}}}&{\frac {i}{\sqrt {2}}}&0\\0&0&1\\\end{pmatrix}}} _{=\;\mathbf {E} }{\begin{pmatrix}e^{i\phi }&0&0\\0&e^{-i\phi }&0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {i}{\sqrt {2}}}&0\\{\frac {1}{\sqrt {2}}}&{\frac {-i}{\sqrt {2}}}&0\\0&0&1\\\end{pmatrix}}$

which gives

$\mathbf {U'} ^{\dagger }\mathbf {R} \mathbf {U'} ={\begin{pmatrix}\cos \phi &-\sin \phi &0\\\sin \phi &\cos \phi &0\\0&0&1\\\end{pmatrix}}\quad {\hbox{with}}\quad \mathbf {U'} =\mathbf {U} {\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {i}{\sqrt {2}}}&0\\{\frac {1}{\sqrt {2}}}&{\frac {-i}{\sqrt {2}}}&0\\0&0&1\\\end{pmatrix}}.$

The columns of U′ are orthonormal. The third column is still n, the other two columns are perpendicular to n. This result implies that any orthogonal matrix R is equivalent to a rotation over an angle φ around an axis n.

Equivalence classes

It is of interest to remark that the trace (sum of diagonal elements) of the real rotation matrix given above is 1 + 2cosφ. Since a trace is invariant under an orthogonal matrix transformation:

$\mathrm {Tr} [\mathbf {A} \mathbf {R} \mathbf {A} ^{\mathrm {T} }]=\mathrm {Tr} [\mathbf {R} \mathbf {A} ^{\mathrm {T} }\mathbf {A} ]=\mathrm {Tr} [\mathbf {R} ]\quad {\hbox{with}}\quad \mathbf {A} ^{\mathrm {T} }=\mathbf {A} ^{-1},$

it follows that all matrices that are equivalent to R by an orthogonal matrix transformation have the same trace. The matrix transformation is clearly an equivalence relation, that is, all equivalent matrices form an equivalence class. In fact, all proper rotation 3×3 rotation matrices form a group, usually denoted by SO(3) (the special orthogonal group in 3 dimensions) and all matrices with the same trace form an equivalence class in this group. Elements of such an equivalence class share their rotation angle, but all rotations are around different axes. If n is a eigenvector of R with eigenvalue 1, then An is an eigenvector of ARA^T, also with eigenvalue 1. Unless A = E, n and An are different.