The Pythagorean Theorem: II. The infinite discrete case

Abstract

The study of the Pythagorean Theorem and variants of it as the basic result of noncommutative, metric, Euclidean Geometry is continued. The emphasis in the present article is the case of infinite discrete dimensionality.

1. Introduction

We continue our study of the Pythagorean Theorem begun in ref. 1. The numbering of results and remarks in ref. 1 will be used in this article in two ways: A reference to Proposition 3 is a reference to that proposition in ref. 1, and the results in this article will be numbered from 13 on (following Proposition 12, the last result in ref. 1).

Our focus in this article is the case of infinite-dimensional Hilbert space and an infinite-dimensional subspace, although we study first the case of a finite-dimensional subspace of infinite-dimensional space. The novelty in that case is that we are projecting the vectors of an infinite orthonormal basis onto the finite-dimensional subspace. Propositions 1 and 2 apply, as they stand, to prove those variants of the Pythagorean Theorem in that situation. The Carpenter's Theorem for that case is quite another matter. Several “proofs” of it were developed, the first of which was invalid: as argued, the “proof” yielded results that did not respect necessary conditions (discovered later). There is a small warning here: the intricacies of the arguments in the case of an infinite-dimensional subspace with infinite-dimensional orthogonal complement are needed to cause certain infinitely repeated processes to produce convergent sums. That convergence is far less automatic than might sometimes seem natural. The shortest of our arguments in the case of the finite-dimensional subspace was not short. Junhao Shen suggested a change in strategy that produces a shortened version, which appears as the proof of Theorem 13. I am happy to express my gratitude for his suggestion. The results involving finite-dimensional subspaces appear in the next section.

In the third section, we study the case of an infinite-dimensional subspace with infinite-dimensional complement. Although the precise formula of Proposition 3 does not apply in that case, the “integrality condition” implicit in that assertion (and mentioned) plays a crucial role in these results.

In the last section, we resume the examination of the relation between the Pythagorean Theorem and doubly stochastic (now, infinite) matrices. We produce such matrices with a block and its complement of finite weight (as promised). Although the formula of Proposition 12 is not applicable here, the integrality condition is and is proved in the concluding section.

2. Finite and Cofinite-Dimensional Subspaces

We prove versions of the Carpenter's Theorem for subspaces of infinite-dimensional Hilbert space ℋ that are finite or cofinite-dimensional in ℋ. From Proposition 2, if we specify numbers in [0, 1], they must have sum m if they are to be the squares of the lengths of the projections of the elements in an orthonormal basis onto some m-dimensional subspace of ℋ. Subject to this condition, is there such a subspace? There is, as we shall show. The proof uses the finite case (Theorem 6) and some additional constructions.

Theorem 13.

If {e_b}_b∈𝔹is an orthonormal basis for the Hilbert space ℋ and numberst_bin [0, 1] are specified, there is an m-dimensional subspace 𝒱 of ℋ such that ∥Fe_b∥² = t_bfor each b in 𝔹, whereF is the projection with range 𝒱, if and only if∑_b∈𝔹t_b = m.

Proof:

Let 𝔹₀ be {b ∈ 𝔹 : t_b = 0} and ℋ₀ the closed linear span of {e_b}_b∈𝔹0. Since ∑_{b∈𝔹∖𝔹0}t_b = m and t_b > 0 when b ∉ 𝔹₀, 𝔹∖𝔹₀is a countable set. Note that m = ∑_{b∈𝔹∖𝔹0}t_b ≤ ∑_{b∈𝔹∖𝔹0} 1 =dim(ℋ⊖ℋ₀). If we find 𝒱 in ℋ⊖ℋ₀ such that ∥Fe_b∥² = t_b for each b in 𝔹∖𝔹₀, we are done, because Fe_b = 0 when b ∈ 𝔹₀. Restricting to ℋ⊖ℋ₀, we may assume that ℋ is separable and that each t_b > 0. We have dealt with the case where ℋ has finite dimension. Henceforth, we assume that ℋ has dimension ℵ₀, that our given orthonormal basis is e₁, e₂, … , and that the specified numbers are a₁, a₂, … .

Some further reductions are useful. By restricting our attention to the orthogonal complement of the subspace of ℋ spanned by the basis elements e_j for which a_j is 0 or 1 and constructing a projection with matrix having diagonal the remaining a_j with respect to the remaining e_j, we may assume that each a_j ∈ (0, 1). Since ∑a_j = m, there are at most a finite number of a_j greater than a given number, and these can be written in decreasing order. Thus, for some permutation π of ℕ, a_π(1) ≥ a_π(2) ≥ ⋯ . Suppose E_π is a projection with matrix relative to e₁, e₂, … having diagonal a_π(1), a_π(2), … . Then U*_πE_πU_π has matrix with diagonal a₁, a₂, … , where U_π is the permutation unitary that maps e_j to e_π⁻¹(j) for each j in ℕ. It suffices, in general, to construct our matrix with the specified diagonal in any order.

We may assume that m ≥ 2 (from Proposition 1) and that a₁ ≥ a₂ ≥ ⋯ with each a_j in (0, 1). Now, ∑a_j = m − a₁ > m − 1. Thus there is an s such that ∑a_j < m − 1 and ∑a_j ≥ m − 1. If t = ∑a_j − (m − 1), then t ≤ a_s+1. From Theorem 6, there is a projection E of rank m − 1 with matrix relative to the basis e₂, e₃, … , e_s+1 having a₂, a₃, … , a_s+1 − t as its diagonal. Moreover, ∑a_j + a₁ + t = 1. Thus, from Proposition 1, there is a projection G of rank 1 with matrix relative to the basis e₁, e_s+2, e_s+3, … whose diagonal is a₁ + t, a_s+2, a_s+3, … . Noting that a_s+1 − t ≤ a_s+1 ≤ a₁ ≤ a₁ + t, we can splice the two projections E and G together and then “permute” the splice, to form a projection F whose matrix relative to the basis e₁, e₂, … has diagonal a₁, a₂, … . ■

Following this fifteenth variation, the situation in which the projections of the elements of our orthonormal basis {e_n} on the subspace 𝒱 of ℋ have lengths whose squares sum to ∞ remains to be studied. In this case, from Proposition 2, 𝒱 must be infinite dimensional. Once again, the question of whether the squares of the lengths can be assigned in [0, 1] arbitrarily, subject only to the condition that their sum diverges is the more involved aspect of this case. Our experience, to this point, makes it tempting to believe that the question has an affirmative answer. Some further consideration makes it clear that there is more to be said.

Again, the question can be formulated in terms of projections, their matrices relative to {e_n}, and diagonals of those matrices. If 𝒱 is the (infinite-dimensional) subspace for the specified lengths a₁, a₂, … and E is the projection with 𝒱 as range, then I − E has ℋ⊖𝒱, the orthogonal complement of 𝒱, as its range. The matrix for the projection I − E has 1 − a₁, 1 − a₂, … as its diagonal. If ∑_j1 − a_j converges, it must converge to an integer, since I − E is a projection. If we choose the a_j such that ∑_j1 − a_j converges, but not to an integer (e.g., let a_j be 1 − (1/j²)), then a_j → 1, whence ∑a_j = ∞, and there is no projection with diagonal a₁, a₂, … relative to {e_j}. (If E were such a projection, I − E would have diagonal 1 − a₁, 1 − a₂, … , with finite sum other than an integer, contradicting Proposition 2). However, if ∑_j1 − a_j = m, with m an integer, then there is a projection I − E with diagonal 1 − a₁, 1 − a₂, … , and hence a projection E with diagonal a₁, a₂, … . This discussion provides our sixteenth variation.

Theorem 14.

If {e_a}_a∈𝔸is an orthonormal basis for the Hilbert space ℋ and{t_a}_a∈𝔸is a family of numbers in [0, 1], there is an infinite-dimensional subspace 𝒱 of ℋ with m-dimensional orthogonal complement such that∥Fe_a∥² = t_afor each a in 𝔸, where F is the projection with range 𝒱, if and only if∑_a∈𝔸1 − t_a = m.

3. Infinite-Dimensional Subspaces with Infinite-Dimensional Complement

In the context of orthonormal bases for Hilbert space, there remains the case where ∑_a∈𝔸t_a and ∑_a∈𝔸1 − t_a diverge. This case leads to our seventeenth variation. As we shall see in the course of the proof, and as noted in the statement, there is more to the story than the divergence of the two sums noted. To recognize this in advance, we need only consider the case where the assigned diagonal entries consist of an infinite number of 0s and terms a₁, a₂, … in [½, 1] such that ∑1 − a_j converges to a number a not an integer. If E is a projection whose matrix has that diagonal, then the restriction of E to the space generated by the basis elements corresponding to all the a_j is also a projection of the sub-Hilbert space onto some subspace of it. Relative to those basis elements corresponding to the a_j, that projection has a matrix whose diagonal has entries a₁, a₂, … . But ∑1 − a_j is a number other than an integer, by assumption. We have seen that such a diagonal is not a possibility for a projection. This restriction appears in more complex form as well. If we replace a finite number of the 0s by numbers r₁, … , r_k in (0, ½) such that (r₁ + ⋯ + r_k) − ∑(1 − a_j) is not an integer, then the resulting assignment of numbers is not the diagonal of a projection. This reduces to the case just considered by noting that, with the present assumption, ∑(1 − a_j) + (1 − r₁) + ⋯ + (1 − r_k) is not an integer. Of course, the restriction on the diagonal in this case is foreshadowed by Proposition 3 (and the remarks following it).

To simplify the discussion and focus on essentials, we deal with the case where ℋ has dimension ℵ₀.

Theorem 15.

Let {e₁, e₂, …} be an orthonormal basis for the Hilbert space ℋ and numbersa₁, a₂, … in [0, 1]be specified. Let a′₁, a′₂, … be the a_jin (½, 1], a"₁, a"₂, … those in[0, ½], a the sum of thea"_j, and b the sum∑1 − a′_j.There is an infinite-dimensional subspace 𝒱 of ℋ with infinite-dimensional complement such that∥Fe_j∥² = a_jfor each j, where F is the projection with range 𝒱, if and only if ∑a_jand ∑1 − a_jdiverge and either of a or b is infinite or both are finite and a − b is an integer.

Proof:

As in the proof of Theorem 13, by restricting to the orthogonal complement of the subspace of ℋ generated by the basis elements e_j for which a_j is either 0 or 1, we may assume that each a_j ∈ (0, 1). Pursuing this same idea, we note that if {ℕ_j}_j=1,2,…is a set of mutually disjoint subsets of ℕ with union ℕ such that we can find a projection E_j with range contained in the closed subspace generated by all e_n for which n ∈ ℕ_j, with ∥E_je_n∥² = a_n, for each such e_n, and E_je_m = 0 for each other e_m, then ∑E_j (= F) is a projection such that ∥Fe_j∥² = a_j for each j.

Let ℕ′ be the set of a_j-indices of a′_j and ℕ" those of a"_j. We suppose, first, that ∑a"_j = ∞. Let n(1) be the least integer n such that a′₁ + a"₁ + ⋯ + a"_n ≥ 3. Since a′₁ < 1 and each a"_j ≤ ½, n(1) ≥ 5. Let b₂, … , b_n(1) be a"₂, … , a"_n(1) rearranged in decreasing order (so, b₂ ≥ b₃ ≥ ⋯ ≥ b_n(1)). Let m(1) be the least integer m such that a′₁ + b₁ + b₂ + ⋯ + b_m ≥ 3, where b₁ = a"₁. Then 5 ≤ m(1) ≤ n(1) and a′₁ + ∑b_j < 3. Let ã be 3 − a′₁ − ∑b_j, b′_m(1)−1 be b_m(1)−1 + ã, and b′_m(1) be b_m(1) − ã. Then a′₁ + ∑b_j + b′_m(1)−1 = 3, and

(∗)

[For the first inequality of (∗), note that a′₁ + ∑b_j ≥ 3, whence b_m(1) ≥ 3 − a′₁ − ∑b_j = ã; for the last inequality, note that b_m(1) is some a"_j (≤ ½) as is b_m(1)−1. Thus b′_m(1)−1 = b_m(1)−1 + ã ≤ b_m(1)−1 + b_m(1) ≤ 1.]

Let ℕ₁ be the set of indices of the a_j in {a′₁, a"₁, … , a"_n(1)} corresponding to a′₁, b₁, … , b_m(1)−1. [Recall that b₁, … , b_n(1) is a rearrangement of a"₁, … , a"_n(1).] Let j(1), j(2), … be the numbers in ℕ"∖ℕ₁ in ascending order, except that j(1) is the index of the a_j in {a"₁, … , a"_n(1)} that b_m(1) represents, and j(2) is the index of the a_j that a′₂ represents. Let n(2) be the least integer n such that a′₂ + b′_m(1) + ∑a_j(k) ≥ 3. Let c₁ be b′_m(1), c₂ be a′₂, and c₃, … , c_n(2) be a_j(3), … , a_j(n(2)) rearranged in decreasing order except that c₃ is a_j(3). [Note that the smallest number in ℕ"∖ℕ₁ is one of j(1) or j(3).] Let m(2) be the least integer m such that ∑c_j ≥ 3. Then ∑c_j < 3, m(2) ≤ n(2), and m(2) ≥ 6. Let b̃ be 3 − ∑c_j, c′_m(2)−1 be c_m(2)−1 + b̃, and c′_m(2) be c_m(2) − b̃. Note that 0 < b̃ ≤ c_m(2) ≤ ½, whence

We repeat this process, letting ℕ₂ be j(1), j(2) and the set of indices of the a_j in {a_j(3), … , a_j(n(2))} corresponding to c₃, … , c_m(2)−1. Let k(1), k(2), … be the numbers in ℕ"∖(ℕ₁ ∪ ℕ₂) in ascending order, except that k(1) is the index of the a_j in {a_j(1), … , a_j(n(2))} that c_m(2) represents, and k(2) is the index of the a_j that a′₃ represents. Let n(3) be the least integer n such that a′₃ + c′_m(2) + ∑a_k(r) ≥ 3. Let d₁ be c′_m(2), d₂ be a′₃, and d₃, … , d_n(3) be a_k(3), … , a_k(n(3)) rearranged in decreasing order except that d₃ is a_k(3). Again, the smallest number in ℕ"∖(ℕ₁ ∪ ℕ₂) is one of k(1) or k(3). Let m(3) be the least integer m such that ∑d_j ≥ 3. Then ∑d_j < 3, m(3) ≤ n(3), and m(3) ≥ 6. Let c̃ be 3 − ∑d_j, d′_m(3)−1 be d_m(3)−1 + c̃, and d′_m(3) be d_m(3) − c̃. Note that 0 < c̃ ≤ d_m(3) ≤ ½ so that

Continuing in this way, we construct disjoint subsets ℕ₁, ℕ₂, … of ℕ with union ℕ. In addition, if r(1), … , r(m(j) − 1) are the elements of ℕ_j, with r(3), … , r(m(j) − 1) in ascending order, there are alterations ã_r(1) and ã_r(m(j)−1) of a_r(1) and a_r(m(j)−1), as described, such that ã_r(1) + a′_j + ∑a_r(k) + ã_r(m(j)−1) = 3. At the same time, with s(1), … , s(m(j + 1) − 1), the elements of ℕ_j+1 and ã_s(1), a_s(2), a_s(3), … , a_{s(m(j+1)−2)}, ã_{s(m(j+1)−1)} summing to 3, with ã_s(1) and ã_{s(m(j+1)−1)} the elements a_s(1) and a_{s(m(j+1)−1)} altered as described, we have ã_r(m(j)−1) + ã_s(1) = a_r(m(j)−1) + a_s(1) and

We include the possibility that ℕ′ is finite (or null) in the foregoing argument. If ℕ′ is {a′₁, … , a′_j−1}, we eliminate the references to “a′_k” in the construction of ℕ_k when k ≥ j.

From Theorem 6, there are three-dimensional projections E_j and E_j+1 whose matrices relative to the bases e_r(1), … , e_r(m(j)−1) and e_s(1), … , e_{s(m(j+1)−1)} have diagonals ã_r(1), a_r(2), a_r(3), … , a_r(m(j)−2), ã_r(m(j)−1) and ã_s(1), a_s(2), a_s(3), … , a_{s(m(j+1)−2)}, ã_{s(m(j+1)−1)}, respectively. We extend each of the projections E_j to a projection (denoted, again, by “E_j”) defined on all of ℋ by letting it annihilate all other basis elements. Then, because the sets ℕ_j are disjoint, E_jE_k = 0 when j ≠ k. Let E be ∑E_j. The next part of this argument is devoted to describing the splicing used to transform E into the projection with the specified diagonal.

We transform E by means of a sequence of unitary operators W_n(θ_n) of the form appearing in the proof of Theorem 7. In the present case, with h(1) the index of the a_j represented by b_m(1)−1 [recall that j(1) is the index of the a_j represented by b_m(1)], W₁(θ₁)e_h(1) =sinθ₁e_h(1) +cosθ₁e_j(1), W₁(θ₁)e_j(1) = −cosθ₁e_h(1) +sinθ₁e_j(1), W₁(θ₁)e_j = e_j for all other e_j, whence

and

Here, θ₁ is chosen [as it may be, by virtue of (∗)] so that the convex combination b′_m(1)−1sin²θ₁ + b′_m(1)cos²θ₁ is b_m(1)−1, from which b′_m(1)−1cos²θ₁ + b′_m(1)sin²θ₁ = b_m(1), since b′_m(1)−1 + b′_m(1) = b_m(1)−1 + b_m(1).

In the same way, we define W₂(θ₂). If g(1) is the index of the a_j represented by c_m(2)−1, then W₂(θ₂)e_g(1) =sinθ₂e_g(1) +cosθ₂e_k(1), W₂(θ₂)e_k(1) =−cosθ₂e_g(1) +sinθ₂e_k(1). [Recall that k(1) is the index of the a_j represented by c_m(2).] Again, W₂(θ₂)EW₂(θ₂)* “splices” c′_m(2)−1 and c′_m(2) to c_m(2)−1 and c_m(2), when θ₂ is suitably chosen, and leaves other “diagonal entries” of E unaltered. Note, too, that W₁(θ₁)W₂(θ₂)e_j = W₂(θ₂)W₁(θ₁)e_j for each j. Here, we use the fact that ℕ₁ and ℕ₂ each contain four or more elements so that no two of j(1), h(1), g(1), and k(1) are equal. Thus W₁(θ₁) and W₂(θ₂) commute.

Note that W_k(θ_k)e_j = e_j unless j ∈ ℕ_k ∪ ℕ_k+1. Thus W_k(θ_k)E_n = E_n if n is neither k nor k + 1, since E_n has range in the space generated by e_j with j in ℕ_n. Suppose j ∈ ℕ_n. Then

In any event, F_r(e_j) = F_s(e_j) when r and s exceed n + 1. Thus {F_r(e_j)} converges as r → ∞ for each fixed j. As {F_r} is a sequence of projections, it is uniformly bounded by 1. The basis {e_j} generates a dense linear manifold on each element of which {F_r} acts to produce a convergent sequence of vectors in ℋ. It follows that {F_rx} is Cauchy convergent, hence convergent for each x in ℋ. If Fx is its limit, then F is linear and ∥F∥ ≤ 1. Thus the sequence of projections {F_r} is strong-operator convergent to F, and F is a projection whose matrix relative to {e_j} has diagonal {a_j} in some order.

The foregoing argument establishes our result when a = ∞. If b = ∞, the argument shows that there is a projection G with b₁, b₂, … as its diagonal, where b_j = 1 − a_j. The diagonal of I − G is a₁, a₂, … (= 1 − b₁, 1 − b₂, …). It remains to treat the case where a and b are finite and a − b is an integer. With this assumption, and the added hypothesis that both the sums of the a_j and of the b_j are infinite, ℕ′ and ℕ" must be infinite sets. There are at most a finite number of a"_j exceeding 1/n for a given positive integer n. By arranging those a"_j in decreasing order and letting n take on the values 1, 2, … , successively, we may re-label the a"_j so that a"₁ ≥ a"₂ ≥ ⋯ . Similarly, we may assume that b"₁ ≥ b"₂ ≥ ⋯ , where b"_j = 1 − a′_j, whence a′₁ ≤ a′₂ ≤ ⋯ . As ∑a"_j and ∑b"_j are finite, we have that a"_j → 0 and b"_j → 0 as j → ∞. It follows that a′_j → 1 as j → ∞. We may assume that a ≤ b (otherwise, we work with b₁, b₂, …).

We discuss a procedure for “distributing the a"_j among a′₁, a′₂, … .” Let n be the smallest integer k such that a"₁ ≤ ∑b"_j. Then ∑b"_j < a"₁. Replace a′₁, … , a′_n−1 by 1, … , 1 and a′_n by a′_n + a"₁ − ∑b"_j(= ã′_n). Then 0 < a′_n < ã′_nand a"₁ − ∑b"_j ≤ b"_n, soã′_n ≤ a′_n + b"_n = 1. In addition,

It follows from Lemma 5 that (a′_n, … , a′₁, a"₁) is contained in the permutation polytope generated by (1, … , 1, ã′_n, 0). As at the end of the proof of Theorem 6, there is a unitary operator U₁ on ℋ such that U₁e_k = e_k for each k not in ℕ₁, the set of indices {j(1), … , j(n + 1)}, of the a_j in {a′_n, … , a′₁, a"₁}, which restricts to 𝔜₁, the subspace of ℋ generated by e_j(1), … , e_j(n+1), as a unitary operator U′₁ such that U′₁AU′₁* has matrix with diagonal a"₁, a′₁, … , a′_n, when A, on 𝔜₁, has matrix with diagonal 0, 1, … , 1, ã′_n relative to the basis e_j(1), … , e_j(n+1).

We now distribute a"₂ among a′_n+1, a′_n+2, … by the procedure just described, forming a finite subset ℕ₂ (= {k(1), … , k(m + 1)}) of ℕ, disjoint from ℕ₁, a unitary operator U₂ on ℋ, such that U₂e_k = e_k for each k not in ℕ₂, whose restriction U′₂ to 𝔜₂, the space generated by e_k(1), … , e_k(m+1), transforms each operator A, on 𝔜₂, whose matrix has diagonal 0, 1, … , 1, ã′_n+m into one whose matrix has diagonal a"₂, a′_n+1, … , a′_n+m relative to the basis e_k(1), … , e_k(m+1).

Continuing in this way, we construct disjoint subsets ℕ₁, ℕ₂, … of ℕ with union ℕ, commuting unitaries U₁, U₂, … on ℋ, and (finite-dimensional) subspaces 𝔜₁, 𝔜₂, … with span dense in ℋ, as described before. Distributing all the a"_j among the a′_j yields an infinite sequence of 0s in place of the a"_j and an infinite sequence ã₁, ã₂, … of numbers in (½, 1] such that ∑1 − ã_j = b − a, an integer, by assumption. (There is “room” for the distribution of all the a"_j among the a′_j from the assumption that a ≤ b.) From Theorem 14, there is a projection E₀ with diagonal ã₁, ã₂, … relative to the basis {e_j}_j∈ℕ′, and of course, a projection E with diagonal 0, 0, … , ã₁, ã₂, … relative to the basis {e_j}. We organize the basis {e_j} according to the sets ℕ₁, ℕ₂, … and the diagonal of the matrix for E such that the entries at the diagonal positions corresponding to numbers in ℕ_jare the numbers obtained from the matching step of the distribution procedure. If we now form U₁EU*₁, U₂U₁EU*₁U*₂, … , successively, we construct a sequence of projections that converges, in the strong-operator topology, to a projection F (by the same argument used for the first part of the proof, where we assumed that a is infinite). The diagonal of F relative to {e_j} is a"₁, a"₂, … , a′₁, a′₂, … , that is, a₁, a₂, … , by choice of U₁, U₂, … .

Having completed the proof of the “Carpenter's Theorem” in this case, we are left with the task of verifying the curious “integrality” condition imposed on a − b. In more detail, we let our orthonormal basis for ℋ be {e_j}_j∈ℤ0, where ℤ₀ is the set of nonzero integers, and F be a projection on ℋ with matrix (a_jk) relative to {e_j}. Let ℤ₋ and ℤ₊ be the negative and positive integers in ℤ₀, respectively. Our assumption now is that ∑a_jj (= a) and ∑1 − a_jj (= b) are finite. We wish to prove that a − b is an integer.

Let E be the projection whose range is spanned by {e_j}_j∈ℤ−. In effect, we have assumed that the positive operators EFE and (I − E)(I − F)(I − E) are of trace class (L₁). It follows that FE, EF, (I − F)(I − E), and (I − E)(I − F) are operators of Hilbert–Schmidt Class (L₂). Thus the sum of all |a_jk|² with j or k in ℤ₋ converges as does the sum of all |b_jk|² with j or k in ℤ₊, where (b_jk) is the matrix of I − F. If T is a Hilbert–Schmidt operator on ℋ, we denote by ∥T∥₂ the Hilbert–Schmidt (L_2⁻) norm of T. That is, ∥T∥ is tr(T*T) (=tr(TT*)), where “tr(T*T)” denotes the sum of the diagonal entries of the matrix for T*T relative to an arbitrary orthonormal basis for ℋ, in particular, relative to {e_j}_j∈ℤ0. Thus ∥T∥ is the sum of the squares of the absolute values of all the entries of the matrix for T (or of T*). Since

with B a bounded operator on ℋ,

and ∥BT∥₂ ≤ ∥B∥∥T∥₂. In this notation, we have ∥EF∥ = ∥FE∥ =tr(EFE) = a and ∥(I − F)(I − E)∥ = b.

Given a positive ɛ (<(2 + 2a^1/2 + 2b^1/2)⁻¹), choose n₀ in ℤ₊ such that ∑_{j or k<−n0}|a_jk|² and ∑_{j or k>n0}|b_jk|² are each less than ɛ′², where ɛ′ = ɛ[28(1 + a^1/2 + b^1/2)]⁻¹. Let A be the matrix that has a_jk as its j,k entry when |j| and |k| do not exceed n₀, 1 at the j,j entry when j > n₀, and 0 at all other entries. Then, by choice of n₀ and A, ∥F − A∥₂ < 2ɛ′. Hence |∥F∥₂ − ∥A∥₂| < 2ɛ′.

Let E₀ be the projection with range spanned by {e₋₁, … , e_−n0}. Again, by choice of n₀ and A, ∥EF − E₀A∥₂ < 2ɛ′. Hence |∥EF∥₂ − ∥E₀A∥₂| < 2ɛ′. Let E′₀ be the projection with range spanned by {e₁, … , e_n0} and I₀ be the projection E₀ + E′₀. Then ∥(I − E)A − (I − E)F∥₂ < 2ɛ′. At the same time, (I − E)(I − A) = E′₀(I − A), whence

Note, too, that

Let A₀ be I₀FI₀. Then A₀ − A = A − A², whence ∥A₀ − A∥₂ < 6ɛ′. Of course, we may treat A₀ as the 2n₀ × 2n₀ matrix (a_jk)_|j|,|k|≤n0. Since A₀ is I₀FI₀, ∥A₀∥ ≤ 1 and A₀ is positive. Let λ_−n0, … , λ_n0 be the 2n₀ eigenvalues of A₀ (with repetitions). Then λ_−n0 − λ, … , λ_n0 − λ are the eigenvalues of A₀ − A. Let ɛ_j be λ_j − λ (= λ_j(1 − λ_j)). If λ_j ≥ ½, then (1 − λ_j)² ≤ 4ɛ. If λ_j ≤ ½, then λ ≤ 4ɛ. We suppose that m₀ of the eigenvalues λ_j are in [½, 1]. Let G₀ be the projection whose matrix relative to the basis that diagonalizes A₀ is diagonal with 1 in place of each of the λ_j in [½, 1] and 0 in place of the other λ_j. Then

and ∥A₀ − G₀∥₂ ≤ 2∥A₀ − A∥₂ < 12ɛ′. Thus ∥E₀A₀ − E₀G₀∥₂ < 12ɛ′. Since E₀A = E₀A₀,

At the same time, (I − E)(I − A) = E′₀(I₀ − A₀), whence

It follows that |∥EF∥₂ − ∥E₀G₀∥₂| < 14ɛ′ < 1, and

From the foregoing, we have that

For the last inequality, we note that ∥E₀G₀∥₂ ≤ 1 + ∥EF∥₂. In the same way,

Thus, as I₀, E₀, and G₀ have ranks 2n₀, n₀, and m₀, respectively, from Proposition 3,

Now, m₀ and n₀ vary with the choice of ɛ, but they are always integers. As a − b is arbitrarily close to an integer, a − b is an integer.

We have established that a − b is an integer for an arbitrary subset 𝒮 of diagonal elements of F with convergent sum a whose complementary set of diagonal elements 𝒮′, subtracted from 1, also has a convergent sum b. To conclude our proof, we note that the existence of any such 𝒮 implies that the set 𝒮₀ of a_jj in [0, ½] is such a set. As 𝒮 has a convergent sum, it contains at most a finite number of a_jj exceeding ½. Similarly, 𝒮′ contains at most a finite number of a_jj less than or equal to ½. Shifting one finite set from 𝒮 to 𝒮′ and the other from 𝒮′ to 𝒮 produces the sets 𝒮₀ and 𝒮′₀ with convergent sums a₀ and b₀, respectively. Moreover, a₀ − b₀ and a − b differ by an integer, as described in the comment following the proof of Proposition 3. Thus a₀ − b₀ is an integer if and only if a − b is. ■

4. Pythagorean Matrices

In Remark 11, we discussed doubly stochastic matrices. We used Proposition 12 to provide another proof of Proposition 3, referring to Proposition 12 as a “Pythagorean Theorem” for finite doubly stochastic matrices. In this section, we consider infinite doubly stochastic matrices. We prove a Pythagorean Theorem for a class of them, the Pythagorean matrices (Proposition 16).

We say that a doubly stochastic matrix A (= (a_jk)) is Pythagorean when there is a Hilbert space ℋ and two orthonormal bases {e_j} and {f_k} for ℋ such that a_jk = |〈e_j, f_k〉|². We show that a Pythagorean Theorem holds for Pythagorean matrices: The difference of the weights of complementary blocks is an integer when those weights are finite. As promised in Remark 11, we establish the existence of doubly stochastic matrices with infinite complementary blocks, each of which has finite weight and no zero diagonal entries. We use the “Carpenter” result in Theorem 15 and the notation of the proof of “integrality” for this.

When j ∈ ℤ₋, let a_j be 2^j. When j ∈ ℤ₊, let a_j be 1 − 2^−j. Since 1 = ∑_j∈ℤ− a_j (= a) and 1 = ∑_j∈ℤ+ 1 − a_j (= b) and a − b = 0, an integer, we may apply Theorem 15 to conclude that there is an infinite-dimensional subspace ℋ₀ of ℋ with infinite-dimensional complement ℋ⊖ℋ₀ (= ℋ′₀) such that ∥Fe_j∥² = a_j, for each j in ℤ₀, where F is the projection of ℋ on ℋ₀. If j ∈ ℤ₊, ∥(I − F)e_j∥² = 1 − ∥Fe_j∥² = 1 − a_j = 2^−j. Let {f_k}_k∈ℤ+ and {f_k}_k∈ℤ− be orthonormal bases for ℋ₀ and ℋ′₀, respectively. Let a_jk be |〈e_j, f_k〉|², so that (a_jk) is an infinite doubly stochastic matrix, as noted before. Since

for each j in ℤ₀, and ∑_k∈ℤ− a_jk = ∥(I − F)e_j∥² = 1 − a_j, for each such j, we have that ∑_j∈ℤ− ∑_k∈ℤ+a_jk = ∑_j∈ℤ−2^j = 1 and ∑_j∈ℤ+∑_k∈ℤ− a_jk = ∑_j∈ℤ+ 2^−j = 1. Thus the weight of each of the ℤ₊, ℤ₋ and ℤ₋, ℤ₊ complementary blocks is 1. Of course, we can construct other examples of infinite complementary blocks, each having finite weight, by using Theorem 15 in the manner just described.

We prove the analogue of Proposition 12 for Pythagorean matrices.

Proposition 16.

The difference of the weights of a block and its complementary block in an infinite Pythagorean matrix is an integer when those weights are finite.

Proof:

Since the complement of a finite block has infinite weight, we may assume that the block A₀ and its complement A′₀ in the matrix A are infinite with finite weights. Assume that A is Pythagorean. We may also assume that A = (a_jk)_j,k∈ℤ0 and that A₀ = (a_jk)_j,k∈ℤ− (so that A′₀ = (a_jk)_j,k∈ℤ+). By assumption, there are orthonormal bases {e_j}_j∈ℤ0 and {f_k}_k∈ℤ0, for a Hilbert space ℋ, such that a_jk = |〈e_j, f_k〉|². Let ℋ₀ be the closure of the space spanned by {f_k}_k∈ℤ− and F the projection of ℋ on ℋ₀. As in the proof of Proposition 12, ∥Fe_j∥² is the sum of the entries of the jth row of A₀, when j ∈ ℤ₋, while ∥(I − F)e_j∥² is the sum of the entries in the jth row of A′₀, when j ∈ ℤ₊. At the same time, ∥Fe_j∥² is the j diagonal entry of the matrix for F and ∥(I − F)e_j∥² is the j diagonal entry for the matrix of I − F (matrices formed relative to the basis {e_j}_j∈ℤ0). Thus w(A₀) is the sum of the j diagonal entries in the matrix for F with j in ℤ₋, and w(A′₀) is the sum of the j diagonal entries in the matrix for I − F (that is, 1 minus the j diagonal entry for F), with j in ℤ₊. These sums are finite, by assumption [because w(A₀) and w(A′₀) are finite]. But Theorem 15 assures us that the difference of these sums is an integer. Thus w(A₀) − w(A′₀) is an integer when A is Pythagorean. ■

There is a great deal more to be said about doubly stochastic matrices in this context, and there are a number of questions that have not been answered (for example: Are there non-Pythagorean doubly stochastic matrices?) That discussion must await another occasion.