 Solution

If $Q(t)$ is a smooth curve of orthogonal matrices with $Q(0)=I$, differentiating $Q(t{)}^{T}Q(t)=I$ gives $Q^{\prime }(0{)}^T+Q^{\prime }(0)=0$. Thus each tangent velocity is a skew-symmetric matrix.

Conversely, if $A^T=-A$, the curve $Q(t)=\exp (tA)$ satisfiesusing commutation of $-tA$ and $tA$. By part (a), $Q^{\prime }(0)=A$. Both inclusions prove