Let $R\in $ SO(3), that is $R$ is real $3\times 3$ orthogonal matrix with determinant $+1$. I am trying prove that if $R= R^\top$, and $R\in $ SO(3) then $R \in \{exp(k\pi \hat{a}) | k\in \mathbb{Z}, a \in S^2 \}$.
Attempt so far : I realize that $R$ will have the following form, $R=QDQ^\top$ where $Q,D\in$ SO(3), and $D$ is diagonal. That is $D\in \{ diag([1\; 1 \;1]), diag([-1 \;-1\; 1]), diag([-1 \; 1 \;-1]), diag([1 \;-1\; -1]) \}$
However, I am unable to see how exponential map comes into the picture.
EDIT : I found another question similar to mine here. One of the answers uses an exponential formula, understanding that probably would help me to prove this.
I think that changing the roation Matrix into a form of a unitary Operator might help.
Starting that an infinitesimal rotation will be the same as a unitary operator, taylor expand both.
A rotation aroundan axis, represented by unit vector a, and an angle $\theta$
$ R( \theta ) = 1 + \frac{\partial R(\theta)}{\partial ( \theta)} \theta \hat{n} + .... $
and a Unitary operator is
$ U(\theta) = 1 + \frac{\partial U(\theta)}{\partial \theta} \theta \hat{n} + ...$
Then, if we introduce a factor k,
$$ \frac{\partial R(\theta)}{\partial \theta} = k \frac{\partial U(\theta)}{\partial \theta} $$
And as any unitary operator can be representated as an exponential,
$$R( \theta) = U(\theta) = e^{{k}{a}\hat{n}}$$
A proof of a rotation having to be around an axis is an assumption and is not proved here.