Sylvster's determinant identity (C.14) is not a consequence of the push-through identity (C.6); use the Schur complement instead

[yousuketakada]

Take the determinants of the two block diagonalizations we use to show the general push-through identity [a generalized version of (C.5)] and the Woodbury identity (C.7). The determinants are both equal to det(M) so that they are equated and setting A and D equal to identities (possibly of different dimensionalities) gives (C.14) with some reparameterization. Note that we cannot take the determinants of both sides of (C.6) and then cancel the det(A) factor because A is not necessarily square nor nonsingular.