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I am trying to compute $e^{A\phi}$ where

$A = \begin{pmatrix} 0 & -x_{3} & x_{2} \\ x_{3} & 0 & -x_{1} \\ -x_{2} & x_{1} & 0 \end{pmatrix} $

and $e^{A\phi}$ is a three dimensional rotation matrix by angle $\phi$ (positive direction)

for a unit vector $\vec{x} = (x_1, x_2, x_3)$.

I currently have

$e^{A\phi} = \sum\limits_{k=0}^\infty \frac{\phi^{k}}{k!}A^k = I + \phi A + \frac{(\phi A)^2}{2!} + \frac{(\phi A)^3}{3!} + \cdots $

I first thought I could use the power series of $sin$ and $cos$

to simplify what I have but it seems like I would need to introduce $i$ for that.

Am I on the right track to compute $e^{A\phi}$ ?

Any help would be great.

Thank you.

Jyrki Lahtonen
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  • $A$ is skew-symmetric and real (I guess). So one eigenvalue must be zero, i.e., $A$ is singular. But a rotation matrix cannot be singular. So, there must be something wrong. – amsmath Oct 24 '19 at 04:30
  • Mathematica gives an utterly horrifying expression for the matrix exponential, so don't expect a nice closed form... – Math1000 Oct 24 '19 at 04:31
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    The matrix is complex normal and therefore complex diagonalizable. Its complex eigenvalues are $0$, $i$, $-i$. Find the corresponding eigenvectors $v_0$, $v_i$, $v_{-i}$ and set up the matrix $U = [v_0,|,v_i,|,v_{-i}]$. Then $A = UDU^$, where $D = diag(0,i,-i)$. So, $e^{\phi A} = \sum_{k=0}^\infty\frac{\phi^k}{k!}UD^kU^ = U,diag(1,e^{i\phi},e^{-i\phi}),U^*$. – amsmath Oct 24 '19 at 04:51
  • It’s $e^{\phi A}$ that’s the rotation matrix, not $A$. – amd Oct 24 '19 at 05:43
  • This is very similar to this old question. – JimmyK4542 Oct 24 '19 at 06:58

2 Answers2

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Assuming your entries are real, you can easily find (using that $x_1^2+x_2^2+x_3^2=1$) that $$ A^3=-A. $$ Then $A^{4}=-A^2$, and \begin{align} e^{tA}&=\sum_{k=0}^\infty\frac{A^{2k}t^{2k}}{(2k)!}+\sum_{k=0}^\infty \frac{A^{2k+1}t^{2k+1}}{(2k+1)!}\\[0.2cm] &=I+A^2\,\sum_{k=1}^\infty\frac{(-1)^{k+1}t^{2k}}{(2k)!}+A\,\sum_{k=0}^\infty \frac{(-1)^kt^{2k+1}}{(2k+1)!}\\[0.2cm] &=I+(1-\cos t)A^2 +(\sin t)\,A. \end{align}

Martin Argerami
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The matrix is complex normal and therefore complex diagonalizable. Its complex eigenvalues are $0$, $i$, $-i$. Find the corresponding eigenvectors $v_0$, $v_i$, $v_{-i}$ and set up the matrix $U = [v_0\,|\,v_i\,|\,v_{-i}]$. Then $A = UDU^*$, where $D = diag(0,i,-i)$. So, $e^{\phi A} = \sum_{k=0}^\infty\frac{\phi^k}{k!}UD^kU^* = U\,diag(1,e^{i\phi},e^{-i\phi})\,U^*$.

Hint for the eigenvectors: The one corresponding to zero is $(x_1,x_2,x_3)^T$. The one I got for $i$ is $(x_1x_3+ix_2,\,ix_1-x_2x_3,\,x_3^2-1)^T$, but this might be wrong due to bad computations of mine.

amsmath
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  • Don't consider the eigenvectors. Use the Lagrange's interpolation polynomial. –  Oct 28 '19 at 11:43
  • @loupblanc For which data points? I don't see which direction you want to point to. – amsmath Oct 28 '19 at 18:20
  • The polynomial $P$ of degree $2$ s.t. $P(0)=1,P(i)=e^{i\phi},P(-i)=e^{-i\phi}$, satisfies $e^{\phi A}=P(A)$. –  Oct 28 '19 at 19:32
  • @loupblanc Ah, nice. That's also possible (and less cumbersome). – amsmath Oct 28 '19 at 20:04