Let \(V\) be \(\RR^3\), with the standard inner product. Let \(W\) be the plane spanned by \(\alpha = (1, 1, 1)\) and \(\beta = (1, 1, -2)\). Let \(U\) be the linear operator defined, geometrically, as follows: \(U\) is rotation through the angle \(\theta\), about the straight line through the origin which is orthogonal to \(W\). There are actually two such rotations—choose one. Find the matrix of \(U\) in the standard ordered basis. (Here is one way you might proceed. Find \(\alpha_1\) and \(\alpha_2\) which form an orthonormal basis for \(W\). Let \(\alpha_3\) be a vector of norm 1 which is orthogonal to \(W\). Find the matrix of \(U\) in the basis \(\{\alpha_1, \alpha_2, \alpha_3\}\). Perform a change of basis.)
Solution
1. The axis of rotation is the line orthogonal to \(W\), spanned by \(\alpha \times \beta\).
the axis is the orthogonal complement of \(W = \operatorname{span}\{\alpha, \beta\}\).
2. \(\alpha \times \beta = (1,1,1) \times (1,1,-2) = (-3, 3, 0)\), with norm \(3\sqrt{2}\), so a unit vector along the axis is \(u = \left(-\frac{1}{\sqrt2}, \frac{1}{\sqrt2}, 0\right)\).
cross product and normalization.
3. By Rodrigues’ rotation formula, the matrix of \(U\) (rotation by \(\theta\) about the unit axis \(u\)) is\[U = \cos\theta\, I + (1 - \cos\theta)\, uu^T + \sin\theta\, [u]_\times,\]where \([u]_\times\) is the cross-product matrix of \(u\).
Rodrigues’ formula.
4. Computing with \(u = \left(-\frac{1}{\sqrt2}, \frac{1}{\sqrt2}, 0\right)\) gives\[U = \begin{pmatrix} \frac{1+\cos\theta}{2} & \frac{\cos\theta - 1}{2} & \frac{\sqrt2}{2}\sin\theta \\[2pt] \frac{\cos\theta - 1}{2} & \frac{1+\cos\theta}{2} & \frac{\sqrt2}{2}\sin\theta \\[2pt] -\frac{\sqrt2}{2}\sin\theta & -\frac{\sqrt2}{2}\sin\theta & \cos\theta \end{pmatrix}.\]
substitute \(u\) into 3 and simplify.
5. Q.E.D.
4.