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→Power series expression: cleaned up ongoing sign errors, showing enough detail to make future debugging easier I hope |
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<math>
\sum_{k=0}^\infty{1 \over k!}B_n^k =\begin{pmatrix}
\sum_{k=0}^\infty{(-1)^k \over 2k!}(\alpha+2\pi n)^{2k} & -\sum_{k=0}^\infty{(-1)^k \over (2k+1)!}(\alpha+2\pi n)^{2k+1} \\
\sum_{k=0}^\infty{(-1)^k \over (2k+1)!}(\alpha+2\pi n)^{2k+1} & \sum_{k=0}^\infty{(-1)^k \over 2k!}(\alpha+2\pi n)^{2k} \\
\end{pmatrix} =\begin{pmatrix}
\cos(\alpha) & -\sin(\alpha) \\
\sin(\alpha) & \cos(\alpha) \\
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Suppose that ''A'' and ''B'' commute, meaning that ''AB'' = ''BA''. Then
: <math>\log{(AB)} = \log{(A)}+\log{(B)}</math>
if and only if <math>\operatorname{arg}(\mu_j) + \operatorname{arg}(\nu_j) \in (- \pi, \pi]</math>, where <math>\mu_j</math> is an [[eigenvalue]] of <math>A</math> and <math>\nu_j</math> is the corresponding [[eigenvalue]] of <math>B</math>.<ref>{{cite journal |last1=APRAHAMIAN |first1=MARY |last2=HIGHAM |first2=NICHOLAS J. |title=The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential |journal=SIAM Journal on Matrix Analysis and Applications |year=2014 |volume=35 |issue=1 |page=97 |doi=10.1137/130920137
: <math> \log{(A^{-1})} = -\log{(A)}.</math>
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== Constraints in the 2 × 2 case ==
If a 2 × 2 real matrix has a negative [[determinant]], it has no real logarithm. Note first that any 2 × 2 real matrix can be considered one of the three types of the complex number ''z'' = ''x'' + ''y'' ''ε'', where
The case where the determinant is negative only arises in a plane with ε<sup>2</sup> =+1, that is a [[split-complex number]] plane. Only one quarter of this plane is the image of the exponential map, so the logarithm is only defined on that quarter (quadrant). The other three quadrants are images of this one under the [[Klein four-group]] generated by ε and −1.
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