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Let A be a $n \times n$ real matrix such that $Exp(A) \in SO(n)$, is it necessarily that $A$ is anti-symmetric?

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Seen this? – J. M. Nov 7 '11 at 16:21
Why is $Exp(tA) \in SO(n)$ for all $t$? – Hezudao Nov 7 '11 at 17:06
up vote 7 down vote accepted

This is false. Let $$ A=\pmatrix{0&\pi\cr-4\pi&0\cr}. $$ Then the eigenvalues of $A$ are $\pm2\pi i$. So $$SAS^{-1}=\pmatrix{0&2\pi\cr-2\pi&0\cr}$$ for an appropriate matrix $S$. But $$ \exp(A)=S^{-1}\exp(SAS^{-1})S=S^{-1}I_2S=I_2 $$ is in $SO(2)$, because for all real $t$ we have $$ \exp\pmatrix{0&t\cr-t&0\cr}=\pmatrix{\cos t&\sin t\cr-\sin t&\cos t\cr}. $$ Yet $A$ is not antisymmetric.

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Notice that the question raised by Adterram about the argument linked to in J.M.'s comment plays a key role. – Jyrki Lahtonen Nov 7 '11 at 19:17

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