 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 givewhere 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 $A$ 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.