Euler's Rotation Theorem - Matrix Proof

Matrix Proof

A spatial rotation is a linear map in one-to-one correspondence with a 3×3 rotation matrix R that transforms a coordinate vector x into X, that is Rx = X. Therefore, another version of Euler's theorem is that for every rotation R, there is a vector n for which Rn = n. The line μn is the rotation axis of R.

A rotation matrix has the fundamental property that its inverse is its transpose, that is

\mathbf{R}^\mathrm{T}\mathbf{R} = \mathbf{R}\mathbf{R}^\mathrm{T} = \mathbf{I},

where I is the 3×3 identity matrix and superscript T indicates the transposed matrix.

Compute the determinant of this relation to find that a rotation matrix has determinant ±1. In particular,

1=\det(\mathbf{I})=\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.

A rotation matrix with determinant +1 is a proper rotation, and one with a negative determinant −1 is an improper rotation, that is a reflection combined with 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. Because this requires that (R − I)n = 0, we see that the vector n must be an eigenvector of the matrix R with eigenvalue λ = 1. Thus, this is equivalent to showing that det(R − I) = 0.

Use the two relations:

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

to compute

\begin{align} \det(\mathbf{R} - \mathbf{I}) =& \det\big((\mathbf{R} - \mathbf{I})^{\mathrm{T}}\big) =\det\big((\mathbf{R}^{\mathrm{T}} - \mathbf{I})\big) = \det\big((\mathbf{R}^{-1} - \mathbf{I})\big) = \det\big(-\mathbf{R}^{-1} (\mathbf{R} - \mathbf{I}) \big) \\ =& - \det(\mathbf{R}^{-1} ) \; \det(\mathbf{R} - \mathbf{I}) = - \det(\mathbf{R} - \mathbf{I})\quad \Longrightarrow\quad \det(\mathbf{R} - \mathbf{I}) = 0. \end{align}

This shows that λ = 1 is a root (solution) of the secular equation, that is,

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

In other words, the matrix R − I 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{I}) \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.

Read more about this topic:  Euler's Rotation Theorem

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