Singular value decomposition for unitary matrices I know the title is strange, but there are many instances in Quantum Information in which one is interested, not in diagonalizing a unitary matrix, but instead in finding its singular value decomposition (SVD from now, the theorem of existence is Cor. 2.4 in Nielsen-Chuang). I was checking how to actually compute this decomposition in here, but the example is based in the usual rectangular matrix $A$ with real entries.
In order to compute $A = USV$, one needs to compute $W = AA^T$ ($T$ is the transpose) and the eigenvectors of $W$ form the columns of $U$; likewise the eigenvectors of $\tilde{W}=A^TA$  form the columns of $V$. Finally $S$ is a diagonal matrix in which the nonzero entries are the squared root of the eigenvalues of $W$.
My question is: how this would be modified if $A$ is unitary. Obviously $AA^\dagger=\mathbb{1}$, so it seems still needed that $W = AA^T$ (is this correct?). What about the diagonal values of $S$? Shouldn't carry a complex conjugation aside from the square root?
 A: In the complex setting, a singular value decomposition is given by a factorization $A = US V^\dagger$ where $U$ and $V$ are unitary matrices of the appropriate sizes and $S$ is a diagonal matrix with nonnegative entries. (Here, $(\cdot)^\dagger$ is the adjoint/conjugate transpose.) If $A$ is unitary, then one singular value decomposition is attained by setting $U = A$ and $S = V = I$ (for $I$ the identity matrix). Another is given by $U = S = I$ and $V = A^\dagger$. Indeed, for any unitary matrix $U$, setting $S = I$ and $V = A^\dagger U$ gives a singular value decomposition of $A$.
The key observation is that all of $A$'s singular values are equal to $1$ (this follows by the fact that $A^\dagger A = AA^\dagger =I$ which has all ones for eigenvalues). The singular vectors are not unique: the singular subspaces associated with a single singular value are unique. As such, for unitary $A$, there is one singular subspace consisting of all vectors, and a singular value decomposition can be constructed by any unitary matrix $U$ whose columns will form a basis for this space.
