Past exam of the mathematical methods I course of the University of Oxford / 2017 / 8 / b / ii

The normalized vectors $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(-1,1{)}^T/\sqrt{2}$ give eigenvalues $I_0$ and $3I_0$. ThereforeThe columns are orthonormal, so $U^{-1}=U^T=U^{\ast }$ and direct multiplication verifies $U^{-1}MU=D$. Signs of individual eigenvectors are free; the displayed signs also make $\det U=+1$.