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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.

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.

Taking the adjoint of $U=M+iN$ gives $U^{†}=M-iN$. In particular, the Hermitian parts of a unitary matrix are uniquely determined:Expanding both unitary identities givesSubtracting shows $MN=NM$, and adding gives $M^2+N^2=I$. If $Mv=\mu v$, commutation implies $M(Nv)=\mu Nv$. The assumed simple spectrum makes the eigenspace one-dimensional, so $Nv=\nu v$. Thus $M,N$ have a common orthonormal eigenbasis.

On such a vector, $Uv=(\mu +i\nu )v$. If the corresponding unitary eigenvalue is $u=e^{i\theta }$, thenThe formulas for $M,N$ also show the same eigenvalue correspondence directly from $Uv=uv$ and $U^{†}v=u^{\ast }v$. Degeneracies do not prevent existence of a common basis, although the simple-spectrum assumption makes the elementary proof immediate.