Editing Matrix exponential (section)

=== Rotation case ===
For a simple rotation in which the perpendicular unit vectors {{Math|'''a'''}} and {{Math|'''b'''}} specify a plane,<ref>in a Euclidean space</ref> the [[Rotation matrix#Exponential map|rotation matrix]] {{mvar|R}} can be expressed in terms of a similar exponential function involving a [[Euler's rotation theorem#Generators of rotations|generator]] {{mvar|G}} and angle {{mvar|θ}}.<ref>{{cite book|last=Weyl|first=Hermann|url=https://books.google.com/books?id=KCgZAQAAIAAJ&pg=PA142 |title=Space Time Matter| year=1952|publisher=Dover|isbn=978-0-486-60267-7|page=142}}</ref><ref>{{cite book|last1=Bjorken|first1=James D.| last2=Drell| first2=Sidney D.|title=Relativistic Quantum Mechanics|url=https://archive.org/details/relativisticquan0000bjor|url-access=registration | publisher=McGraw-Hill|year=1964|page=[https://archive.org/details/relativisticquan0000bjor/page/22 22]}}</ref>
<math display="block">\begin{align}
  G &= \mathbf{ba}^\mathsf{T} - \mathbf{ab}^\mathsf{T} & P &= -G^2 = \mathbf{aa}^\mathsf{T} + \mathbf{bb}^\mathsf{T} \\
  P^2 &= P & PG &= G = GP ~,
\end{align}</math>
<math display="block">\begin{align}
  R\left( \theta \right) = e^{G\theta}
    &= I + G\sin (\theta) + G^2(1 - \cos(\theta)) \\
    &= I - P + P\cos (\theta) + G\sin (\theta ) ~.\\
\end{align}</math>

The formula for the exponential results from reducing the powers of {{mvar|G}} in the series expansion and identifying the respective series coefficients of {{math|''G<sup>2</sup>''}} and {{mvar|G}} with {{math|−cos(''θ'')}} and {{math|sin(''θ'')}} respectively. The second expression here for {{math|''e<sup>Gθ</sup>''}} is the same as the expression for {{math|''R''(''θ'')}} in the article containing the derivation of the [[Euler's rotation theorem#Generators of rotations|generator]], {{math|1=''R''(''θ'') = ''e<sup>Gθ</sup>''}}.

In two dimensions, if <math>a = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]</math> and <math>b = \left[ \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right]</math>, then <math>G = \left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix} \right]</math>, <math>G^2 = \left[ \begin{smallmatrix}-1 & 0 \\ 0 & -1\end{smallmatrix} \right]</math>, and
<math display="block">R(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{bmatrix} = I \cos(\theta) + G \sin(\theta)</math>
reduces to the standard matrix for a plane rotation.

The matrix {{math|1=''P'' = −''G''<sup>2</sup>}} [[Projection (linear algebra)|projects]] a vector onto the {{math|ab}}-plane and the rotation only affects this part of the vector. An example illustrating this is a rotation of {{math|1=30° = π/6}} in the plane spanned by {{math|'''a'''}} and {{math|'''b'''}},

<math display="block">\begin{align}
  \mathbf{a} &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} &
  \mathbf{b} &= \frac{1}{\sqrt{5}}\begin{bmatrix} 0 \\ 1 \\ 2 \\ \end{bmatrix}
\end{align}</math>
<math display="block">\begin{align}
  G = \frac{1}{\sqrt{5}}&\begin{bmatrix}
     0 & -1 & -2 \\
     1 &  0 &  0 \\
     2 &  0 &  0 \\
  \end{bmatrix} &
  P = -G^2 &= \frac{1}{5}\begin{bmatrix}
     5 & 0 & 0 \\
     0 & 1 & 2 \\
     0 & 2 & 4 \\
  \end{bmatrix} \\
  P\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix} =
    \frac{1}{5}&\begin{bmatrix} 5 \\ 8 \\ 16 \\ \end{bmatrix} =
    \mathbf{a} + \frac{8}{\sqrt{5}}\mathbf{b} &
  R\left(\frac{\pi}{6}\right) &= \frac{1}{10}\begin{bmatrix}
     5\sqrt{3} &      -\sqrt{5} &     -2\sqrt{5} \\
      \sqrt{5} &   8 + \sqrt{3} & -4 + 2\sqrt{3} \\
     2\sqrt{5} & -4 + 2\sqrt{3} &  2 + 4\sqrt{3} \\
  \end{bmatrix} \\
\end{align}</math>

Let {{math|1=''N'' = ''I'' - ''P''}}, so {{math|1=''N''<sup>2</sup> = ''N''}} and its products with {{math|''P''}} and {{math|''G''}} are zero. This will allow us to evaluate powers of {{math|''R''}}.

<math display="block">\begin{align}
       R\left( \frac{\pi}{6} \right) &= N + P\frac{\sqrt{3}}{2} + G\frac{1}{2} \\
     R\left( \frac{\pi}{6} \right)^2 &= N + P\frac{1}{2} + G\frac{\sqrt{3}}{2} \\
     R\left( \frac{\pi}{6} \right)^3 &= N + G \\
     R\left( \frac{\pi}{6} \right)^6 &= N - P \\
  R\left( \frac{\pi}{6} \right)^{12} &= N + P = I \\
\end{align}</math>

{{further|Rodrigues' rotation formula|Axis–angle representation#Exponential map from so(3) to SO(3)}}