The claim
(a) Minimal realizations. Let k : S × S → F, F = ℝ or ℂ, be positive definite. For i = 1, 2 let Φᵢ : S → Hᵢ satisfy k(s,t) = ⟨Φᵢ(s), Φᵢ(t)⟩, with the Φᵢ(s) spanning a dense subspace of Hᵢ. Then there is exactly one unitary U : H₁ → H₂ with UΦ₁(s) = Φ₂(s) for every s. Over ℝ it is an orthogonal isomorphism.
(b) Finite matrices. Let X, Y ∈ ℝ^(n×d), whose rows are two representations of the same n inputs. If XXᵀ = YYᵀ, then Y = XQ for some Q ∈ O(d).
(c) Centred. Let C = I − (1/n)11ᵀ. If CXXᵀC = CYYᵀC, then Y = XQ + 1bᵀ for some Q ∈ O(d) and b ∈ ℝ
The argument
(a) This is NOTE-829’s argument, the book’s proof of Thm 4.35 with π_i(A)u_i replaced by Φᵢ(s).
- Define U on finite combinations by U Σ c_s Φ₁(s) := Σ c_s Φ₂(s).
- For both i, ‖Σ c_s Φᵢ(s)‖² = Σ_(s,t) c̄_s c_t k(s,t), the same number.
- So a combination that vanishes in H₁ also vanishes in H₂. Hence U is well defined, linear and isometric on span Φ₁(S).
- An isometry on a dense subspace extends uniquely to an isometry on H₁. Its range is complete, hence closed. It contains span Φ₂(S), which is dense in H₂, so the range is H₂ and U is unitary.
- U is unique because any unitary with UΦ₁(s) = Φ₂(s) is fixed on the dense set span Φ₁(S).
Nothing here uses complex scalars, so over ℝ it gives an orthogonal isomorphism.
(b)
- Let xⱼ and yⱼ be the rows. Apply (a) with S = {1, …, n}, k(j,l) = (XXᵀ)_(jl), H₁ = row space of X and H₂ = row space of Y. This gives an orthogonal isomorphism U : H₁ → H₂ with Uxⱼ = yⱼ.
- dim H₁ = rank X = rank XXᵀ = rank YYᵀ = dim H₂ = r.
- The orthogonal complements H₁^⊥ and H₂^⊥ in ℝ^d therefore both have dimension d − r. Choose any orthogonal isomorphism V between them.
- Then Q′ = U ⊕ V is orthogonal on ℝ^d with Q′xⱼ = yⱼ for every j. Q = Q′ᵀ gives Y = XQ.
When r < d, V is arbitrary, so Q is not unique.
(c) Apply (b) to CX and CY, whose Gram matrices are equal by hypothesis. Then CY = CXQ. Write Y = CY + 1ȳᵀ and X = CX + 1x̄ᵀ, where x̄ and ȳ are the column means. Then Y = XQ + 1(ȳ − Qᵀx̄)ᵀ.
Where it is used
- THEORY-004. Its clause that a representation is determined by its centred kernel up to a translation and an orthogonal map is (c). The promotion history (version 6) took Cor. 1.35’s confirmation as the record’s own. It also flagged that representation matrices whose rows do not span ℝ^d need the isometry extended across the complement. That is step (b)3.
- LIT-250 §5.3 states (b) as known, without proof.
What it does not say
- Nothing about kernels estimated from data, or about approximate equality. If CXXᵀC and CYYᵀC are only close, (c) says nothing; that is what the Procrustes–Bures identity in THEORY-004 measures.
- Nothing about realizations that are not minimal. In infinite dimensions, a non-minimal realization gives U only as an isometry onto a closed subspace.