 Solution

The real orthogonal group and special orthogonal group are, under matrix multiplication,An orthogonal matrix preserves Euclidean inner products; taking determinants of $Q^{T}Q=I$ shows that its determinant is $1$ or $-1$. Its inverse is $Q^T$, so $Q{Q}^T=I$ as well.

For $Q\in SO(3)$, use $Q-I=Q(I-Q^T)$. The determinant then satisfiesIt follows that $\det (Q-I)=0$. Thus the real matrix $Q-I$ has a nonzero vector in its kernel of a linear map, yieldingThis is the three-dimensional case of odd-dimensional special orthogonal transformation has a fixed vector; geometrically it supplies the fixed rotation axis, unless the transformation is the identity and every direction is fixed.

The claim is false for $O(3)$. The matrix $Q=-I_3$ is orthogonal with determinant $-1$ and has only the eigenvalue $-1$. It therefore has no nonzero eigenvector with eigenvalue $1$.