Computing exponential of matrix with power series

Question

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.

Answer 1

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}

Answer 2

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.