Solution
= Solution
If $Q(t)$ is a smooth curve of <orthogonal matrices> with $Q(0)=I$, differentiating $Q(t)^TQ(t)=I$ gives $Q'(0)^T+Q'(0)=0$. Thus each tangent velocity is a <skew-symmetric matrix>.
Conversely, if $A^T=-A$, the curve $Q(t)=\exp(tA)$ satisfies
$$
Q(t)^TQ(t)=\exp(-tA)\exp(tA)=I,
$$
using commutation of $-tA$ and $tA$. By part (a), $Q'(0)=A$. Both inclusions prove
$$
\boxed{T_IO(n)=\{A:A^T=-A\}=\mathcal A}.
$$