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<record version="14" id="6786">
 <title>rotation matrix</title>
 <name>RotationMatrix</name>
 <created>2005-02-19 10:02:40</created>
 <modified>2006-06-13 08:18:38</modified>
 <type>Definition</type>
 <creator id="1858" name="matte"/>
 <author id="13753" name="Mathprof"/>
 <author id="3771" name="CWoo"/>
 <author id="1858" name="matte"/>
 <classification>
	<category scheme="msc" code="15-00"/>
 </classification>
 <synonyms>
	<synonym concept="rotation matrix" alias="rotational matrix"/>
 </synonyms>
 <related>
	<object name="OrthogonalMatrices"/>
	<object name="ExampleOfRotationMatrix"/>
	<object name="DecompositionOfOrthogonalOperatorsAsRotationsAndReflections"/>
	<object name="DerivationOfRotationMatrixUsingPolarCoordinates"/>
	<object name="DerivationOf2DReflectionMatrix"/>
	<object name="TransitionToSkewAngledCoordinates"/>
 </related>
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 <content>\begin{defn} A \emph{rotation matrix} is a 
    (real) orthogonal matrix whose determinant is $+1$. 
All $n\times n$ rotation matrices form a group called
    the \emph{specia{l} orthogona{l} grou{p}} and it is denoted by 
    $\operatorname{SO}(n)$. 
\end{defn}

\subsubsection*{Examples}
\begin{enumerate}
\item The identity matrix in $\R^n$ is a rotation matrix.
\item The most general rotation matrix in $\R^2$ can be written as 
$$
  \begin{pmatrix} 
       \cos \theta &amp; -\sin \theta \\  
      \sin \theta &amp; \cos \theta
  \end{pmatrix},
$$
where $\theta\in \R$. 
Multiplication (from the left) with this matrix 
rotates a vector (in $\R^2$) $\theta$ radians in the anti-clockwise
direction. 
\end{enumerate}

\subsubsection*{Properties}
\begin{enumerate}
\item Suppose $v\in \R^n$ is a unit vector. 
  Then there exists a rotation matrix $R$
  such that $R\cdot v = (1,0,\ldots, 0)$.
\item 
In fact, for $v\in \R^n$, $n\ge 3$, there are many rotation matrices 
$\mathbf{R} \in \operatorname{SO}(n)$ such that 
$R\cdot v = (1,0,\ldots, 0)^T$.
To see this, let $f$ be the mapping   
  $f\colon \operatorname{SO}(n-1)\rightarrow \operatorname{SO}(n)$,
  defined as
$$
 f(Q)=
 \begin{pmatrix}
 1 &amp; 0_{1\times n-1}\\
 0_{n-1\times 1} &amp; Q_{n-1\times n-1}
 \end{pmatrix}.
$$
Then for each $Q\in \operatorname{SO}(n-1)$, $f(Q)$ 
    maps $(1,0,\ldots, 0)^T$ onto itself. Thus, if 
    $R_0 \in \operatorname{SO}(n)$ satisfies $R\cdot v=(1,0,\ldots, 0)^T$, 
    then $f(Q)\cdot R$ satisfies the same property for all 
    $Q\in \operatorname{SO}(n-1)$. 
\end{enumerate}</content>
</record>
