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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.

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