# Prove the logarithm of a matrix diverges

I am trying to prove that $$\log{B}$$ does not converge for any matrix $$B$$ in the set $$X = \theta(3)-SO(3)$$. (where $$\theta(n)$$ is $$n\times n$$ orthogonal matrices and $$SO(n)$$ is orthogonal matrices with determinant $$1$$.)

We know that $$B.B^t=I$$ and $$\det(B) \neq 1$$ since $$B\in X$$

This implies $$\fbox{det(B) = -1}$$ since,

$$\det(B.B^t)=\det(B).\det(B^t)=\det(B)^2=\det(I)=1$$

I tried assuming that $$\log{B}$$ does converge and finding a contradiction.

Then, $$\log{B^t}$$ also converges and

$$\log{B} + \log{B^t} = \log{(B.B^t)} = \log(I) = 0$$

Taking the determinant of both sides,

$$\det\log(B) = \det(-\log(B^t)) = - \det\log(B)$$ since $$B$$ is $$odd\times odd$$

Then, we get, $$\fbox{det(log(B))=0}$$

I have no idea where I can go from here or whether these findings are true or not.

Any bits of help is appreciated.