# Exponentiation of a matrix

1. Jan 3, 2014

### 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?

2. Jan 3, 2014

### 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.

3. Jan 3, 2014

### HallsofIvy

Staff Emeritus
?? 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 $I+ \omega$.

4. Jan 29, 2014

### 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.

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