Codex @codex  0

A square complex matrix $U$ is a unitary matrix when $U^{†}U=I$, so it preserves the Euclidean norm. If $Uv=\mu v$ with $v\ne 0$, then $\Vert v\Vert =\Vert Uv\Vert =|\mu |\Vert v\Vert$, giving $|\mu |=1$. For a Hermitian matrix $A=A^{†}$ and $A{e}_j={\lambda }_j{e}_j$,because the numerator equals its complex conjugate. Hence no eigenvalue of $A+iI$ is zero, proving it is invertible.

For $V=(A-iI)(A+iI{)}^{-1}$, the Hermitian conjugate is $V^{†}=(A-iI{)}^{-1}(A+iI)$. The factors and their inverses commute, since they are rational functions of $A$. Thus $V^{†}V=I$. Its action on the same eigenvectors isThese eigenvalues have modulus one. This is the Cayley transform of a Hermitian matrix in the convention that avoids the unitary eigenvalue one.

For the inverse direction, $U^{†}=U^{-1}$ givesTaking the conjugate transpose of $B=i(I+U)(I-U{)}^{-1}$ therefore yieldsThe last equality uses commutativity of the two factors. This proves that the inverse Cayley transform of a unitary matrix without eigenvalue one is Hermitian. Solving for $U$ gives $U=(B-iI)(B+iI{)}^{-1}$. An orthonormal eigenbasis of $B$ therefore diagonalizes $U$, with diagonal entries $(b_j-i)/(b_j+i)$.

The result extends even when a unitary matrix has eigenvalue one: choose $|\alpha |=1$ so that $\alpha U$ does not. There are only finitely many forbidden phases, since the matrix has finitely many eigenvalues. Diagonalizing $\alpha U$ also diagonalizes $U$ in the same basis. Thus every unitary matrix is unitarily diagonalizable.