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

Famous quotes containing the words matrix and/or proof:

    “As all historians know, the past is a great darkness, and filled with echoes. Voices may reach us from it; but what they say to us is imbued with the obscurity of the matrix out of which they come; and try as we may, we cannot always decipher them precisely in the clearer light of our day.”
    —Margaret Atwood (b. 1939)

    “Sculpture and painting are very justly called liberal arts; a lively and strong imagination, together with a just observation, being absolutely necessary to excel in either; which, in my opinion, is by no means the case of music, though called a liberal art, and now in Italy placed even above the other two—a proof of the decline of that country.”
    —Philip Dormer Stanhope, 4th Earl Chesterfield (1694–1773)