Proof of Theorem 2, p. 47 — for B* = Diag(1/√n_k)O with O orthogonal, ‖b*_l − b*_k‖ = √(1/n_k + 1/n_l)

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Proof of Theorem 2, p. 47 — for B* = Diag(1/√n_k)O with O orthogonal, ‖b*_l − b*_k‖ = √(1/n_k + 1/n_l) · Prove2Me