How Does Matrix Exponentiation Retrieve Rotation Matrices?

[gentsagree]

I need to retrieve a finite rotation matrix (with cos and sin) from the exponentiation of the infinitesimal version of it.

Suppose my infinitesimal matrix is ω. I then compute exp(ω).

My guess would be

\exp(\omega)=\sum_{k=0}\frac{\omega^{2k}}{2k!}+\sum_{k=0}\frac{\omega^{2k+1}}{(2k+1)!}

i.e. the even and odd contributions.

The notes I'm reading suggest instead:

\exp(\omega)=I+\sum_{k=1}\frac{\omega^{2k}}{2k!}+\sum_{k=1}\frac{\omega^{2k+1}}{(2k+1)!}

which looks weird to me; if I take the identity matrix I to be the k=0 contribution of the even part (ω^0=1), then I don't know where the term linear in ω is in the series any more. I think it's not there at all.

Even more: I do need the k=0 contributions later on to retrieve the series expansion expressions for cos and sin.

What do you think? Any comments?

 

[AlephZero]

I agree. Replacing the $\omega^0/0!$ term by $I$ is fair enough, but the second sum should be from $k = 0$.

It's probably just a typo.

 

[HallsofIvy]

?? The "linear term in \omega" is the term with \omega^1 which means it is the term in the second sum with k= 0: \frac{\omega^{2(0)+ 1}}{(2(0)+ 1)!}= \omega.

The linear approximation to e^\omega is I+ \omega.

 

[FactChecker]

I agree. It's a typo. If you are doing this, don't forget that you can use the characteristic equation of the matrix ω to eliminate all the high powers.