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I have a skew symmetric matrix $$C=\left( \begin{array}{ccc} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{array} \right).$$.

Question

1) How do I compute $e^{C}$ with out series expansion of $e^{C}$ . Means shall I get a finite expression for it?

NB : Values of $a_i$ s cant be changed

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Let $x = \sqrt{a_1^2+a_2^2+a_3^2}$. You can verify that $C^3 = -(a_1^2+a_2^2+a_3^2)C = -x^2C$.

Hence, $C^{2m+1} = (-1)^mx^{2m}C$ and $C^{2m} = (-1)^{m-1}x^{2m-2}C^2$.

Therefore, $e^C = \displaystyle\sum_{n=0}^{\infty}\dfrac{1}{n!}C^n = I + \sum_{m=0}^{\infty}\dfrac{1}{(2m+1)!}C^{2m+1} + \sum_{m=1}^{\infty}\dfrac{1}{(2m)!}C^{2m}$

$= \displaystyle I + \sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m}}{(2m+1)!}C + \sum_{m=1}^{\infty}\dfrac{(-1)^{m-1}x^{2m-2}}{(2m)!}C^{2}$

$= \displaystyle I + \dfrac{1}{x}\sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m+1}}{(2m+1)!}C - \dfrac{1}{x^2}\sum_{m=1}^{\infty}\dfrac{(-1)^{m}x^{2m}}{(2m)!}C^{2}$

$= I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$

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  • $\begingroup$ Thanks... May I ask you one question is this $e^x$ is applicable to matrices of 3d and more $\endgroup$ – Nirvana Jul 27 '14 at 7:16
  • $\begingroup$ For any square matrix $A$, we define $e^A = \sum_{n=0}^{\infty} \frac{1}{n!}A^n$. $\endgroup$ – JimmyK4542 Jul 27 '14 at 7:18
  • $\begingroup$ That is fine..Agreed...I have a case where matrices of 3D and 4D involved..Will that satisfy that dimension too..? $\endgroup$ – Nirvana Jul 27 '14 at 7:19
  • $\begingroup$ What do you mean by matrices of 3D and 4D? Are you talking about tensors? $\endgroup$ – JimmyK4542 Jul 27 '14 at 7:21
  • $\begingroup$ yes I meant tensors $\endgroup$ – Nirvana Jul 27 '14 at 7:23
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A geometric proof.

$C$ is the matrix of $u\in \mathbb{R}^3\rightarrow a\times u$ where $a=[a_1,a_2,a_3]^T$. $\exp(C){\exp(C)}^T=I$ and $\det(\exp(C))>0$ ; then $\exp(C)\in O^+(3)$, that is a rotation $\exp(C)=R=Rot(\Delta,\theta)$. Note that $Ca=0$ implies $Ra=a$ and $\Delta$ is the oriented line defined by $a$.

After some calculations, $Trace(R)=1+2\cos(\theta)=1+2\cos(||a||)$. Moreover, let $b=[a_2,-a_1,0]^T$ ($b$ is orthogonal to $a$) ; we obtain $\dfrac{1}{{a_1}^2+{a_2}^2}||b\times Rb||=\sin(||a||)$. Conclusion $\theta=||a||$.

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  • $\begingroup$ What is $u$ and $a$ here? and is $\|u\|= 1$? $\endgroup$ – sleeve chen Feb 18 '17 at 1:11

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