= Solution
Use the <orthogonal matrix> identity $A(I-A^T)=A-I$. The product rule for <determinants>, $\det A=1$, and invariance of determinant under <matrix transpose> give
$$
\det(A-I)=\det A\det(I-A^T)=\det(I-A)=-\det(A-I),
$$
where the last sign comes from dimension three. Hence $\det(A-I)=0$. The real <linear map> $A-I$ has a nonzero real vector in its <kernel>, so \b[$A$ has an eigenvector with eigenvalue one]. This is the <odd-dimensional special orthogonal transformation has a fixed vector> property, here proved directly without first classifying the other eigenvalues.
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