Unitary matrix

A complex matrix is unitary when $U^{†}U=U{U}^{†}=I$. Its columns form an orthonormal basis, and it preserves inner products and norms.

Suppose a unitary matrix $U$ is symmetric under ordinary transpose, and write its entrywise real and imaginary parts as $A$ and $B$. They are real symmetric matrices. Expanding $U^{†}U=I$ gives $A^2+B^2+i(AB-BA)=I$, proving the displayed identities. Commuting real symmetric matrices admit simultaneous diagonalization: diagonalize $A$, then diagonalize the restriction of $B$ to each invariant eigenspace of $A$. A common real orthonormal eigenbasis givesIf $A$ has simple eigenvalues, every eigenvector of $A$ is already an eigenvector of $B$, because its one-dimensional eigenspace is invariant. The eigenvalues of $A$ and $B$ are paired as the real and imaginary parts of the same eigenvalue of $U$.

The displayed parts of a unitary matrix are skew-Hermitian matrices. Expanding $U{U}^{†}=U^{†}U=I$ gives $AB=BA$ and $A^2+B^2=-I$. They admit a common orthonormal eigenbasis. For a unitary eigenvalue $u$, the corresponding part eigenvalues are $i\mathrm{Im}u$ and $-i\mathrm{Re}u$. If $U^T=-U$, then $U^{†}=-\overline{U}$, so these are exactly its real and imaginary parts and both are real skew-symmetric matrices. Such an invertible antisymmetric matrix has even dimension, by its determinant. If one part has distinct eigenvalues, commutation alone shows its one-dimensional eigenspaces are preserved by the other part.

Every unitary matrix has a unique decomposition $U=A+iB$ into Hermitian matrices. The formulas follow by adding and subtracting $U$ and its conjugate transpose. Expanding $U^{†}U=U{U}^{†}=I$ gives $A^2+B^2+i(AB-BA)=I$ and $A^2+B^2-i(AB-BA)=I$. Therefore the parts commute and $A^2+B^2=I$.