# Angle-Axis Parametrization of SO(3) Proof

Suppose we have an element $$R$$ of $$SO(3)$$. $$R$$ is characterized by, $$R^T = R^{-1}$$. There are a number of equivalent characterizations such as $$R$$ preserves norms or dot products.

I am looking for a succinct proof that any $$R$$ can be written as

$$R = e^{i\theta \hat{n}\cdot J}$$

Here $$\hat{n}$$ is a unit vector and $$J$$ is a a vector of the $$J$$ matrices which are the generators of 3D rotations (members of the Lie Algebra for $$SO(3)$$: $$\mathfrak{so}(3)$$). I would like a clean proof of this statement. I think I have some of the pieces of the proof..

## Start of proof?:

Because $$R$$ is unitary its eigenvalues of modulus $$1$$. They also come in complex conjugate pairs. Since $$SO(3)$$ has dimension $$3$$ this means one of the eigenvalues is purely real and the other two are complex conjugate pair. The real eigenvalue is either $$\pm 1$$ but since we are considering $$SO(3)$$ we have $$\det(R)=+1$$ which implies that the real eigenvalue is $$+1$$. Thus all elements of $$SO(3)$$ have $$+1$$ as an eigenvalue. Call the eigenvector of $$R$$ with eigenvalue $$+1$$ by $$\hat{n}$$ so that

$$R\hat{n} = \hat{n}$$

The other two eigenvalues can be written as $$e^{\pm i\theta}$$ since they have modulus one and come in a complex conjugate pair.

Let $$L = e^{i\theta \hat{n} \cdot J}$$

We must prove that $$L=R$$. Let $$A = i\theta \hat{n}\cdot J$$. Since $$J$$ is skew-symmetric, $$J^T=-J$$ we have that $$A$$ is also skew-symmetric so that $$A^T=-A$$. Since the $$A$$ matrices are all skew symmetric it is the case that

$$L^T L = e^{A^T}e^A = e^{-A}e^A = 1$$

This means that $$L$$ is a rotation.

Next, it can be proven that $$(\hat{n}\cdot J)v = \hat{n}\times v$$ for any $$v$$. This means $$Av = i\theta \hat{n} \times v$$. In particular $$A\hat{n} = i\theta \hat{n}\times \hat{n} =0$$. This means $$\hat{n}$$ is an eigenvector of $$A$$ with eigenvalue $$0$$. This means that $$\hat{n}$$ is also an eigenvector of $$L=e^A$$ with eigenvalue $$1=e^0$$.

Next it can be seen by direct computation (and the fact that $$|\hat{n}|^2=1$$) that the eigenvalues of $$\hat{n}\cdot J$$ are $$\{0,i,-i\}$$ so that the eigenvalues of $$A=i\theta \hat{n}\cdot J$$ are $$\{1,e^{i\theta}, e^{-i\theta}\}$$.

This is where I'm finally stumped. I see that $$R$$ and $$L$$ share all of their eigenvalues and at least 1 eigenvector-eigenvalue pair. I also know that all of the eigenvectors are orthogonal so that the remaining two eigenvalues for each (which are complex conjugate pairs) lie in the plane. I lack some characterization of these other two eigenvectors to complete the proof that the matrices $$R$$ and $$L$$ share eigenvalues and eigenvectors (and thus are equal).

What is the final ingrediant I need to complete this proof?

## A more general approach?

I can prove that generally if a unitary matrix can be written as $$U=e^{iHt}$$ (with $$t$$ real) that $$H$$ must be Hermitian or that the combination $$iHt$$ must be anti-Hermitian. The proof I know for this is related to taking the derivative of $$U$$ with respect to $$t$$ and then applying the unitary property $$U^{\dagger}=U^{-1}$$. It can in fact be seen above that the $$A$$ matrix is hermitian. Here $$t$$ seems to be related to $$\theta$$. In any case, if I could prove generally that any unitary matrix can be written as $$e^{iHt}$$ for some hermitian matrix $$H$$ and real number $$t$$ then it would not be too bad to show $$R=L$$. Basically the proof would say that since $$R$$ is a unitary matrix it can be written as $$e^{iHt}$$ with $$H$$ hermitian. Since $$R$$ is real it could be proven that $$H$$ is real (and thus anti-symmetric). The space of real anti-symmetric matrices is spanned by the $$J$$ matrices so it would follow that $$iHt$$ must be equal to $$i\theta \hat{n}\cdot J$$ for some $$\theta$$ and $$\hat{n}$$.

Is this more general approach valid? How to prove that any unitary matrix can be written as $$U=e^{A}$$ with $$A$$ anti-hermitian?

Any answer to my two bolded questions at the bottom of each section would be much appreciated.