Past exam of the mathematical methods I course of the University of Oxford / 2015 / 12 / vi / Solution

With only ${\alpha }_3$ nonzero, $T=\mathrm{diag}({\alpha }_3,-{\alpha }_3)$ andFor ${\alpha }_3\notin 2\pi \mathbb{Z}$, the eigenvectors are the nonzero multiples of $(1,0{)}^T$ and $(0,1{)}^T$. When ${\alpha }_3=2\pi k$, $U=(-1{)}^{k}I$ and every nonzero vector is an eigenvector.