# For which orthogonal $2\times2$ matrices does the exponential become orthogonal?

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.

## 1 Answer

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.

• 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. – Keith Aug 14 at 14:23
• 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. – John Hughes Aug 14 at 14:26
• @Keith See here if you want to write a matrix using MathJax. – Thomas Shelby Aug 14 at 14:27
• @JohnHughes OK, thanks to your advice, I wrote those matrices; it seems easier than I thought. Now, are they correct? – Keith Aug 14 at 14:40