Let A be a $n \times n$ real matrix such that $Exp(A) \in SO(n)$, is it necessarily that $A$ is anti-symmetric?
The exponential of a real matrix
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linear-algebra
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2Why is $Exp(tA) \in SO(n)$ for all $t$? – 2011-11-07
1 Answers
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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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1Notice that the question raised by Hezudao about the argument linked to in J.M.'s comment plays a key role. – 2011-11-07