Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 6B (Vectors and Matrices)

The characteristic polynomial of the displayed rotation matrix is $(1-\lambda )({\lambda }^2-2\lambda \cos \theta +1)$, so its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, all of modulus one. The real eigenvector $(0,0,1{)}^T$ spans the rotation axis, while the arguments $\pm \theta$ of the conjugate eigenvalues record the rotation angle on its perpendicular plane.

Applying first the $z$-rotation and then the $x$-rotation givesIts $1$-eigenspace is spanned by $(1,-1,1{)}^T$, so $\mathbf{n}=(1,-1,1{)}^T/\sqrt{3}$. The matrix trace identity $\mathrm{tr}R=1+2\cos \varphi$ gives $0=1+2\cos \varphi$. Thus $\varphi =2\pi /3$.

The quadratic form has symmetric matrixwhose characteristic polynomial is $(\lambda -2)(\lambda -3)(\lambda -8)$. All its eigenvalues are positive, so $A$ is positive definite and ${\mathbf{x}}^{T}A\mathbf{x}=1$ is an ellipsoid. The spectral theorem for real symmetric matrices gives semi-axis lengthsThe nearest points lie in the eigendirection of the largest eigenvalue. Since $(5,2,1{)}^T$ is an $8$-eigenvector, they are