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

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