Rotation matrix

A rotation of a 3-dimensional rigid body is a motion of the body that leaves one point, O, fixed. By Euler's theorem follows that then not only the point is fixed but also an axis—the rotation axis— through the fixed point. Write ${\hat {n}}$ for the unit vector along the rotation axis and φ for the angle over which the body is rotated, then the rotation is written as ${\mathcal {R}}(\varphi ,{\hat {n}}).$

Erect three Cartesian coordinate axes with the origin in the fixed point O and take unit vectors ${\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}$ along the axes, then the rotation matrix $\mathbf {R} (\varphi ,{\hat {n}})$ is defined by its elements $R_{ji}(\varphi ,{\hat {n}})$:

${\mathcal {R}}(\varphi ,{\hat {n}})({\hat {e}}_{i})=\sum _{j=x,y,x}{\hat {e}}_{j}R_{ji}(\varphi ,{\hat {n}})\quad {\hbox{for}}\quad i=x,y,z.$

In a more condensed notation this equation is written as

${\mathcal {R}}(\varphi ,{\hat {n}})\left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right)=\left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right)\;\mathbf {R} (\varphi ,{\hat {n}}).$

Given a basis of a linear space, the association between a linear map and its matrix is one-to-one.

Properties of matrix

Since rotation conserves the shape of a rigid body, it leaves angles and distances invariant. In other words, for any pair of vectors ${\vec {a}}$ and ${\vec {b}}$ in $\mathbb {R} ^{3}$ the inner product is invariant,

$\left({\mathcal {R}}({\vec {a}}),\;{\mathcal {R}}({\vec {b}})\right)=\left({\vec {a}},\;{\vec {b}}\right).$

A linear map with this property is called orthogonal. It is easily shown that a similar vector/matrix relation holds. First we define

${\vec {a}}=\left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right){\begin{pmatrix}a_{x}\\a_{y}\\a_{z}\end{pmatrix}}\equiv \left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right)\mathbf {a} \quad {\hbox{and}}\quad {\vec {b}}=\left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right){\begin{pmatrix}b_{x}\\b_{y}\\b_{z}\end{pmatrix}}\equiv \left({\hat {e}}_{x},\;{\hat {e}}_{y},\;{\hat {e}}_{z}\right)\mathbf {b}$

and observe that the inner product becomes by virtue of the orthonormality of the basis vectors

$\left({\vec {a}},\;{\vec {b}}\right)=\mathbf {a} ^{\mathrm {T} }\mathbf {b} \equiv \left(a_{x},\;a_{y},\;a_{z}\right){\begin{pmatrix}b_{x}\\b_{y}\\b_{z}\end{pmatrix}}\equiv a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}.$

The invariance of the inner product under ${\mathcal {R}}$ leads to

${\big (}\mathbf {R} \mathbf {a} {\big )}^{\mathrm {T} }\;\mathbf {R} \mathbf {b} =\mathbf {a} ^{\mathrm {T} }\mathbf {R} ^{\mathrm {T} }\;\mathbf {R} \mathbf {b}$

since this holds for any pair a and b it follows that a rotation matrix satisfies

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

where E is the 3×3 identity matrix. For finite-dimensional matrices one shows easily

$\mathbf {R} ^{\mathrm {T} }\mathbf {R} =\mathbf {E} \quad \Longleftrightarrow \quad \mathbf {R} \mathbf {R} ^{\mathrm {T} }=\mathbf {E} .$

A matrix with this property is also called orthogonal. Writing out the two matrix products it follows that both the rows and the columns of the matrix are orthonormal (normalized and orthogonal). Indeed,

${\begin{aligned}\mathbf {R} ^{\mathrm {T} }\mathbf {R} &=\mathbf {E} \quad \Longrightarrow \quad \sum _{k=1}^{3}R_{ki}\,R_{kj}=\delta _{ij}\quad {\hbox{(columns)}}\\\mathbf {R} \mathbf {R} ^{\mathrm {T} }&=\mathbf {E} \quad \Longrightarrow \quad \sum _{k=1}^{3}R_{ik}\,R_{jk}=\delta _{ij}\quad {\hbox{(rows)}}\\\end{aligned}}$

where δ_ij is the Kronecker delta.

Orthogonal matrices come in two flavors: proper (det = 1) and improper (det = −1) rotations. Invoking some properties of determinants, one can prove

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

Compact notation

A compact way of presenting the same results is the following. Designate the columns of R by r₁, r₂, r₃, i.e.,

$\mathbf {R} =\left(\mathbf {r} _{1},\,\mathbf {r} _{2},\,\mathbf {r} _{3}\right)$.

The matrix R is orthogonal if

$\mathbf {r} _{i}\cdot \mathbf {r} _{j}=\delta _{ij},\quad i,j=1,2,3.$

The matrix R is a proper rotation matrix, if it is orthogonal and if r₁, r₂, r₃ form a right-handed set, i.e.,

$\mathbf {r} _{i}\times \mathbf {r} _{j}=\sum _{k=1}^{3}\,\varepsilon _{ijk}\mathbf {r} _{k}.$

Here the symbol × indicates a cross product and $\varepsilon _{ijk}$ is the antisymmetric Levi-Civita symbol,

${\begin{aligned}\varepsilon _{123}=&\;\varepsilon _{312}=\varepsilon _{231}=1\\\varepsilon _{213}=&\;\varepsilon _{321}=\varepsilon _{132}=-1\end{aligned}}$

and $\varepsilon _{ijk}=0$ if two or more indices are equal.

The matrix R is an improper rotation matrix if its column vectors form a left-handed set, i.e.,

$\mathbf {r} _{i}\times \mathbf {r} _{j}=-\sum _{k=1}^{3}\,\varepsilon _{ijk}\mathbf {r} _{k}\;.$

The last two equations can be condensed into one equation

$\mathbf {r} _{i}\times \mathbf {r} _{j}=\det(\mathbf {R} )\sum _{k=1}^{3}\;\varepsilon _{ijk}\mathbf {r} _{k}$

by virtue of the the fact that the determinant of a proper rotation matrix is 1 and of an improper rotation −1. This was proved above, an alternative proof is the following: The determinant of a 3×3 matrix with column vectors a, b, and c can be written as scalar triple product

$\det \left(\mathbf {a} ,\,\mathbf {b} ,\,\mathbf {c} \right)=\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )$.

It was just shown that for a proper rotation the columns of R are orthonormal and satisfy,

$\mathbf {r} _{1}\cdot (\mathbf {r} _{2}\times \mathbf {r} _{3})=\mathbf {r} _{1}\cdot \left(\sum _{k=1}^{3}\,\varepsilon _{23k}\,\mathbf {r} _{k}\right)=\varepsilon _{231}=1.$

Likewise the determinant is −1 for an improper rotation.

Explicit expression

Let ${\overrightarrow {OP}}\equiv {\vec {r}}$ be a vector pointing from the fixed point O of a rotating rigid body to an arbitrary point P of the body. A rotation of this arbitrary vector around the unit vector ${\hat {n}}$ over an angle φ can be written as

${\mathcal {R}}(\varphi ,{\hat {n}})({\vec {r}}\,)=\left[{\vec {r}}-({\hat {n}}\cdot {\vec {r}}\,)\;{\hat {n}}\right]\cos \varphi +({\hat {n}}\times {\vec {r}}\,)\sin \varphi .$

where • indicates an inner product and the symbol × a cross product.