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.


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