# Is a real logarithm of a special orthogonal matrix necessarily skew-symmetric?

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?

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 $-I_2$.

• 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. Feb 15, 2016 at 17:31
• @Matta Thanks for the correction, I've updated the answer. Feb 16, 2016 at 10:01