\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Orthogonal transformation, Proper Orthogonal Transformation, \(\m{SO}(3)\)

Definition : An orthogonal transformation \(\u{A}\) in \(\m{R}^N\) is a second-order tensor that preserves the lengths and the angles between vectors.

If \(\u{A}\) is an orthogonal transformation, we say that \(\u{A} \in \m{O}(N)\) and its special property is that if \(\u{A}\) and \(\u{B} \in \m{O}(N)\) then \(AB \in \m{O}(N)\).

(i) From the definition, it follows that \(\u{A}^{\t{T}} = \u{A}^{-1}\). See proof below.

(ii) In fact the condition \(\u{A}^{\t{T}} = \u{A}^{-1}\) is a sufficient condition for \(\u{A} \in \m{O}(N)\)

(iii) \(\u{A}^{\t{T}} = \u{A}^{-1}\) \(\Rightarrow\) \(\t{det}(\u{A})=\pm 1\) (Proof follows this section)

(i). Proper Orthonormal transformation

An \(\u{A} \in \m{O}(N)\) is called a proper orthogonal transformation if \(\t{det}(\u{A})=+1\).

(ii). Special Orthonormal group

The subset of all proper orthonormal transformations is called the special orthonormal group \(\m{SO}(3)\).

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Proofs

1) Proof that an orthonormal transformation \(\u{A}\) satisfies the equation \(\u{A}^{\t{T}} = \u{A}^{-1}\)

Let \(\u{x}^{\prime} = \u{A}\u{x}\) and \(\u{y}^{\prime} = \u{A}\u{y}\). Since angles are unchanged

\[\begin{align} \frac{\u{x}^{\prime}\cdot\u{y}^{\prime}}{\left\|\u{x}^{\prime}\right\| \left\|\u{y}^{\prime}\right\|} = \frac{\u{x}^\cdot\u{y}}{\left\|\u{x}\right\| \left\|\u{y}\right\|} \end{align}\]

Since lengths are unchanged \(\left\|\u{x}^{\prime}\right\| = \left\|\u{x}\right\|\) and \(\left\|\u{y}^{\prime}\right\| = \left\|\u{y}\right\|\). Therefore

\[\begin{align} \u{x}^{\prime}\cdot\u{y}^{\prime} &= \u{x}^\cdot\u{y} \\ (\u{A}\u{x})^{\t{T}}(\u{A}\u{y}) &= \u{x}^{\t{T}}\u{y} \\ \u{x}^{\t{T}}\u{A}^{\t{T}}\u{A}\u{y} &= \u{x}^{\t{T}}\u{I}\u{y} \\ \u{x}^{\t{T}}(\u{A}^{\t{T}}\u{A}-\u{I})\u{y} &= 0 \quad \forall \quad \u{x}, \u{y} \end{align}\]

In indicial notation,

\[\begin{align} (a^{\t{T}}_{ik}a_{kj}-\delta_{ij})x_{i}y_{i} \end{align}\]

Through various choices of \(\u{x}\) and \(\u{y}\), we can show that

\[\begin{align} & a^{\t{T}}_{1k}a_{k1}-\delta_{11} = 0, a^{\t{T}}_{1k}a_{k3}-\delta_{13} = 0 \\ & a^{\t{T}}_{2k}a_{k2}-\delta_{22} = 0, a^{\t{T}}_{3k}a_{k1}-\delta_{31} = 0 \\ & a^{\t{T}}_{3k}a_{k3}-\delta_{33} = 0, a^{\t{T}}_{2k}a_{k3}-\delta_{23} = 0 \\ & a^{\t{T}}_{1k}a_{k2}-\delta_{12} = 0, a^{\t{T}}_{3k}a_{k2}-\delta_{32} = 0 \\ & a^{\t{T}}_{2k}a_{k1}-\delta_{21} = 0 \end{align}\]

i.e;

\[\begin{align} a^{\t{T}}_{ik}a_{kj}=\delta_{ij} \Rightarrow \u{A}^{\t{T}}\u{A} = \u{I} \Rightarrow \u{A}^{\t{T}} = \u{A}^{-1} \end{align}\]

2) Proof that \(\t{det}(\u{A}) = \pm 1\) if \(\u{A}^{\t{T}} = \u{A}^{-1}\)

\[\begin{align} \t{det}(\u{A}\u{A}^{-1}) &= \t{det}(\u{I}) = 1 \\ \t{det}(\u{A}\u{A}^{\t{T}}) &= 1 \\ \t{det}(\u{A})\t{det}(\u{A}^{\t{T}}) &= 1 \quad (\t{det}(\u{A})=\t{det}(\u{A}^{\t{T}}),\, \textrm{see proof below}) \\ \t{det}(\u{A})^2 &= 1 \\ \t{det}(\u{A}) &= \pm 1 \end{align}\]

3) Proof that \(\t{det}(\u{A})=\t{det}(\u{A}^{\t{T}})\)

\[\begin{align} \t{det}(\u{A}) = & A_{11} (A_{22}A_{33}-A_{32}A_{23}) \\ - & A_{21} (A_{12}A_{33}-A_{32}A_{13}) \\ & A_{31} (A_{12}A_{23}-A_{22}A_{13}) \end{align}\] \[\begin{align} \t{det}(\u{A}^{\t{T}}) = & A_{11} (A_{22}A_{33}-A_{23}A_{32}) \\ - & A_{12} (A_{21}A_{33}-A_{23}A_{31}) \\ & A_{13} (A_{21}A_{32}-A_{22}A_{31}) \end{align}\]

If you rearrange the terms, we see that \(\t{det}(\u{A})=\t{det}(\u{A}^{\t{T}})\).

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