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The exponential map from the Lie algebra of skew-symmetric matrices $\mathfrak{so}(n)$ to the Lie group $\operatorname{SO}(n)$ is surjective and so I know that given any special orthogonal matrix there exists a skew-symmetric real logarithm.

However, must all real logarithms of a special orthogonal matrix be skew-symmetric?

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1 Answer 1

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No. Let's say that $A$ is real a logarithm of $-I_2 \in \operatorname{SO}(2)$. Then for any invertible $P$ we have

$$ e^{PAP^{-1}} = P e^A P^{-1} = P (-I_2) P^{-1} = -I_2 $$

so $PAP^{-1}$ is also a real logarithm of $-I_2$. If you start with a skew-symmetric logarithm of $-I_2$ and conjugate it by a general invertible matrix, there is no reason that you'll get a skew-symmetric matrix. For example, if

$$ A = \begin{pmatrix} 0 & \pi \\ -\pi & 0 \end{pmatrix}, P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$

then

$$ PAP^{-1} = \begin{pmatrix} -\pi & 2\pi \\ -\pi & \pi \end{pmatrix} $$

is a real non-skew-symmetric logarithm of $-I_2$.

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  • $\begingroup$ Except the sneaky -1 in the corner of the last matrix. I think it should be 1. Might be wrong, certainly pedantic. Thank you for the answer. $\endgroup$
    – Matta
    Feb 15, 2016 at 17:31
  • $\begingroup$ @Matta Thanks for the correction, I've updated the answer. $\endgroup$
    – levap
    Feb 16, 2016 at 10:01

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