 Solution

Consider $F(A)=A^{T}A$ with values in the symmetric matrices. At the identity,which is onto the $n(n+1)/2$-dimensional space of symmetric matrices. The same rank holds at every orthogonal matrix. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the special orthogonal group, has dimensionThe tangent equation at the identity is $X^T+X=0$, so