I'm trying to show that the Lie Algebra for $O(n)$ is the set of $n \times n$ skew-symmetric matrices. Here is what I have so far.
Since $O(n)$ is the union of two disjoint subsets, the matrices with determinant of $1$ ($SO(n)$) and determinant $-1$, then the Lie Algebra of $O(n)$ is the same as the Lie Algebra for $SO(n)$ because of course, $I \in SO(n)$. I'll denote the set of skew-symmetric matrices by $Sk(n)$ (is there a convention for denoting this set?) and the Lie Algebra of $O(n)$ with $\mathfrak{so}(n)$.
A skew-symmetric matrix is such that $-M = M^\top$. Then $\exp(-M) = [\exp(M)]^{-1}$. So then, $\exp(M^\top) = [\exp(M)]^{-1}$. But the LHS equals $[\exp(M)]^\top$. Thus, $\exp(m) [\exp(M)]^\top = I$ which implies that $\exp(M) \in O(n)$. Consider $A \in Sk(n)$. We know that $\exp(A) \in O(n)$. In fact, it is not hard to show that $\exp(tA) \in O(n)$ as well. Consider the curve $\alpha(t) = \exp(tA)$. Now, $\alpha(0) = I$, $\alpha'(t) = A \exp(tA)$, and $\alpha'(0) = A$. So we've shown that $Sk(n) \subseteq \mathfrak{so}(n)$.
However, I'm having trouble showing the other direction: $\mathfrak{so}(n) \subseteq SK(n)$. I don't think this direction should require much theory beyond the definitions. Hints are very welcome.
Thank you.
Best Answer
Well let's see. Say $$\gamma:(-1,1)\to O(n)$$is differentiable and $\gamma(0)=I$; we need to show that $$\gamma'(0)+\gamma'(0)^T=0.$$
We know that $\gamma(t)\gamma(t)^T=I$. Differentiating that gives $$\gamma'(0)\gamma(0)^T+\gamma(0)\gamma'(0)^T=0.$$Oh. We're done, because $\gamma(0)=I$.