Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 5B (Algebra and Geometry)

A positive, counterclockwise rotation takes the standard basis vectors $(1,0{)}^T$ and $(0,1{)}^T$ to $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. These images are the columns of its rotation matrix, soIt fixes the origin, preserves the dot product and has determinant one.

Since $A{A}^T=I$ and $\det A=1$, $A$ is an orthogonal matrix in the special orthogonal group. It also satisfies $A^{T}A=I$ and thus preserves the dot product. Normalize the fixed eigenvector to $e_3=v/\Vert v\Vert$. For $u\cdot e_3=0$,so the orthogonal complement of the fixed axis is invariant. Choose an oriented orthonormal basis $e_1,e_2,e_3$. In this basis the matrix has form $\mathrm{diag}(B,1)$, where $B$ is a real $2\times 2$ orthogonal matrix of determinant one.

Write the first column of $B$ as $(\cos \theta ,\sin \theta {)}^T$. Its unit second column must be perpendicular to the first; positivity of the determinant selects $(-\sin \theta ,\cos \theta {)}^T$. Thus $B=R(\theta )$, and $A$ fixes the axis through $v$ while rotating its perpendicular plane by $\theta$. Since trace is invariant under change of basis,This proves that $A$ is the required axial rotation. The trace determines the unsigned angle; the action in the oriented perpendicular plane determines its sign. The identity case has angle zero and any axis.

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.