Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 2A (Vectors and Matrices)

Taking the conjugate transpose of $U=A+iB$ and using that both parts are Hermitian matrices gives $U^{†}=A-iB$. Addition and subtraction recover the Hermitian parts of a unitary matrix:These depend only on $U$, so any such decomposition has the same $A,B$ and is unique. Conversely the displayed expressions are Hermitian matrices and sum to $U$ in the form $A+iB$, establishing existence as well as uniqueness. Their commutation and quadratic relation are proved in the two labelled parts.

By the unitary matrix identities,Subtracting the two equations yields $2i(AB-BA)=0$. ThereforeThe subtraction is important: entrywise real and imaginary parts cannot generally be used to separate these terms, since the Hermitian matrices need not have real entries.

Adding the two expanded unitary matrix identities instead cancels the commutator and gives $2(A^2+B^2)=2I$. HenceTogether with part (i), this characterizes the commuting Hermitian parts of a unitary matrix.