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Let $ A$ be a $2\times2$ real orthogonal matrix. Then when does $e^A$ become orthogonal as well? According to my calculations, $A$ must be skew-symmetric also and therefore there are only two possibilities. Am I correct?

Edit~~~ Ok, the two matrices are $\pmatrix{0 &1 \\ -1 &0}$ and $\pmatrix{0 & -1 \\ 1 &0}$. Thank you all for the advice below. Now, are these matrices correct? I think they are, but I would like others to check in case I am missing something.

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That does indeed sound correct. (And you can probably generalize fairly easily to higher dimensions.)

As a general principle, it's worth knowing that the tangent space to the orthogonal group at $I$ consists exactly of skew-symmetric matrices; since $A \mapsto \exp A$ sends this tangent space to the orthogonal group (and indeed, ignoring speed for the moment, $t \mapsto \exp(tA)$ is a geodesic through the identity, for any skew-symmetric $A$), there's a pretty strong correspondence between the two.

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  • $\begingroup$ Could you pleae show me the two matrices? I actually am not familiar with writing matrices on the latex so...I would like to check my answers are indeed correct. $\endgroup$ – Keith Aug 14 at 14:23
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    $\begingroup$ To write the identity matrix, you type \pmatrix{1 & 0 \\ 0 & 1} and get $\pmatrix{1 & 0 \\ 0 & 1}$; more generally, \pmatrix{a & b \\ c & d} gives $\pmatrix{a & b \\ c & d}$. To answer your question: no, I won't write out the two matrices for you. That's a good exercise for you. $\endgroup$ – John Hughes Aug 14 at 14:26
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    $\begingroup$ @Keith See here if you want to write a matrix using MathJax. $\endgroup$ – Thomas Shelby Aug 14 at 14:27
  • $\begingroup$ @JohnHughes OK, thanks to your advice, I wrote those matrices; it seems easier than I thought. Now, are they correct? $\endgroup$ – Keith Aug 14 at 14:40

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