Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 8B / c / Solution

First note that if $A$ is normal, thenfor every $v$. Hence an eigenvector $v$ with $Av=\lambda v$ also satisfiesNormalize $v$ and extend it to an orthonormal basis. Let $U$ have these basis vectors as columns. The first column of $U^{†}AU$ is $(\lambda ,0,\dots ,0{)}^T$. For every $w\perp v$,so the first row also has no off-diagonal entries. ThereforeComparing the two block products in the normality identity shows that $B{B}^{†}=B^{†}B$, so $B$ is normal.

The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the eigenvector $v$ gives an orthonormal eigenbasis for $A$. This proves the unitary diagonalization of a normal matrix.