 Solution

Let the columns of the given unitary matrix be $u_j$, with $M{u}_j={\lambda }_j{u}_j$. Columnwise, $MU=UD$ for $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Multiplication by $U^{\ast }$ and unitarity giveThe premise includes an orthonormal eigenbasis. It is not true that an arbitrary matrix admits such a basis; over $\mathbb{C}$ it exists precisely for a normal matrix. The real symmetric inertia matrix in part (b) does satisfy the required condition.