The Kadison-Singer and Paulsen Problems in Finite Frame Theory

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6 Peter G. Casazza(2) The square sum of the coefficients of every row equals 1.The row vectors of the matrix B are not (δ, 2)-Rieszable, for any δ independentof N.Proof. A direct calculation yields (1) and (2).We will now show that the column vectors of B are not uniformly twoRieszable independent of N. So let {A 1 , A 2 } be a partition of {1, 2, . . . , 4N}.Without loss of generality, we may assume that |A 1 ∩ {1, 2, . . . , 2N}| ≥ N.Let the column vectors of the matrix B be {ϕ i } 4Ni=1 as elements of C2N . LetP N−1 be the orthogonal projection of C 2N onto the first N − 1 coordinates.Since |A 1 | ≥ N, there are scalars {a i } i∈A1 so that ∑ i∈A 1|a i | 2 = 1 andP N−1( ∑i∈A 1a i ϕ i)= 0.Also, let {ψ j } 2Nj=1 be the orthonormal basis consisting of the original columnsof the DF T 2N . We now have:‖ ∑a i ϕ i ‖ 2 = ‖(I − P N−1 )( ∑a i ϕ i )‖ 2i∈A 1 i∈A 1= 2N + 1 ‖(I − P N−1)( ∑≤ 2N + 1 ‖ ∑= 2N + 1= 2N + 1 .a i ψ i ‖ 2i∈A 1∑|a i | 2i∈A 1i∈A 1a i ψ i )‖ 2Letting N → ∞, this class of matrices is not (δ, 2)-Rieszable, and hence not(δ, 2)-pavable for any δ > 0.If this argument could be generalized to yield non-(δ, 3)-Rieszable (Pavable)matrices, then such an argument will lead to a complete counterexample toPC.1.2.2 The R ɛ -ConjectureIn this section we will define the R ɛ -Conjecture and show it is equivalent tothe Paving Conjecture.Definition 3. A family of vectors {ϕ i } M i=1 is an ɛ-Riesz basic sequence for0 < ɛ < 1 if for all scalars {a i } M i=1 we have
1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 7M∑M∑M∑(1 − ɛ) |a i | 2 ≤ ‖ a i ϕ i ‖ 2 ≤ (1 + ɛ) |a i | 2 .i=1i=1A natural question is whether we can improve the Riesz basis bounds fora unit norm Riesz basic sequence by partitioning the sequence into subsets.Conjecture 2 (R ɛ -Conjecture). For every ɛ > 0, every unit norm Riesz basicsequence is a finite union of ɛ-Riesz basic sequences.This conjecture was first stated by Casazza and Vershynin and was firststudied in [16] where it was shown that PC implies the conjecture. Oneadvantage of the R ɛ -Conjecture is that it can be shown to students at thebeginning of a course in Hilbert spaces.The R ɛ -Conjecture has a natural finite dimensional form.Conjecture 3. For every ɛ > 0 and every T ∈ B(l N 2 ) with ‖T e i ‖ = 1 for i =1, 2, . . . , N there is an r = r(ɛ, ‖T ‖) and a partition {A j } r j=1 of {1, 2, . . . , N}so that for all j = 1, 2, . . . , r and all scalars {a i } i∈Aj we have(1 − ɛ) ∑i∈A j|a i | 2 ≤ ‖ ∑i=1i∈A ja i T e i ‖ 2 ≤ (1 + ɛ) ∑i∈A j|a i | 2 .Now we show that the R ɛ -Conjecture is equivalent to PC.Theorem 2. The following are equivalent:(1) The Paving Conjecture.(2) For 0 < ɛ < 1, there is an r = r(ɛ, B) so that for every N ∈ N, ifT : l N 2 → l N 2 is a bounded linear operator with ‖T ‖ ≤ B and ‖T e i ‖ = 1 forall i = 1, 2, . . . , N, then there is a partition {A j } r j=1 of {1, 2, . . . , N} so thatfor each 1 ≤ j ≤ r, {T e i } i∈Aj is an ɛ-Riesz basic sequence.(3) The R ɛ -Conjecture.Proof. (1) ⇒ (2): Fix 0 < ɛ < 1. Given T as in (2), let S = T ∗ T . Since S hasones on its diagonal, by the Paving Conjecture there is a r = r(ɛ, ‖T ‖) and apartition {A j } r j=1 of {1, 2, . . . , N} so that for every j = 1, 2, . . . , r we havewhere δ =‖ ∑‖Q Aj (I − S)Q Aj ‖ ≤ δ‖I − S‖ɛ‖S‖+1 . Now, for all x = ∑ Ni=1 a ie i and all j = 1, 2, . . . , r we havei∈A ja i T e i ‖ 2 = ‖T Q Aj x‖ 2 = 〈T Q Aj x, T Q Aj x〉 = 〈T ∗ T Q Aj x, Q Aj x〉= 〈Q Aj x, Q Aj x〉 − 〈Q Aj (I − S)Q Aj x, Q Aj x〉≥ ‖Q Aj x‖ 2 − δ‖I − S‖‖Q Aj x‖ 2≥ (1 − ɛ)‖Q Aj x‖ 2 = (1 − ɛ) ∑i∈A j|a i | 2 .
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