= Solution
The real <orthogonal group> and <special orthogonal group> are, under <matrix> multiplication,
$$
\boxed{O(n)=\{Q\in M_n(\mathbb R):Q^TQ=I\},\qquad
SO(n)=\{Q\in O(n):\det Q=1\}.}
$$
An <orthogonal matrix> preserves Euclidean <inner products>; taking <determinants> of $Q^TQ=I$ shows that its <determinant> is $1$ or $-1$. Its inverse is $Q^T$, so $QQ^T=I$ as well.
For $Q\in SO(3)$, use $Q-I=Q(I-Q^T)$. The <determinant> then satisfies
$$
\det(Q-I)=\det Q\det(I-Q^T)=\det(I-Q)=(-1)^3\det(Q-I).
$$
It follows that $\det(Q-I)=0$. Thus the real <matrix> $Q-I$ has a nonzero <vector> in its <kernel of a linear map>, yielding
$$
\boxed{Qv=v\quad\text{for some }v\ne0.}
$$
This 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 \b[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> $1$.
Back to article page
