In linear algebra terms, this can be justified as follows. Any g of U(n) that leaves | ψ ψ ⟩ ⟩ invariant must have | ψ ψ ⟩ ⟩ as an eigenvector. Since the corresponding eigenvalue must be a complex number of modulus 1, this gives the U(1) factor of the isotropy group. The other part of the isotropy group is parametrized by the unitary matrices on the orthogonal complement of | ψ ψ ⟩ ⟩ , which is isomorphic to U(n − 1). From this the assertion of the theorem follows from basic facts about transitive group actions of compact groups.