The Kadison-Singer and Paulsen Problems in Finite Frame Theory

14 Peter G. Casazza∑|〈x, ϕ i 〉| 2 > B − 1.Hence,It follows that for all j,‖Q j P (T x)‖ 2 ≥i∈A j∩{1,2,...,M}∑i∈A j|〈x, ψ i 〉| 2 > 1 − 1 B .O∑|〈Q j P (T x), e i 〉| 2 = ∑i=1= ∑i∈A j|〈x, ψ i 〉| 2 > 1 − 1 B .i∈A j|〈T x, e i 〉| 2Thus, ‖Q j P Q j ‖ = ‖Q j P ‖ 2 > 1 − 1 B. Now, the matrix A = P − D has zerodiagonal and satisfies ‖A‖ ≤ 1 + 1 Band the above shows that for any K × Kdiagonal projections {Q j } r j=1 with ∑ rj=1 Q j = Id we have‖Q j AQ j ‖ ≥ ‖Q j P Q j ‖ − ‖Q j DQ j ‖ ≥ 1 − 2 , for some j .BFinally, as B = r → ∞, we obtain a sequence of examples which negate thePaving Conjecture.Weaver [43] also shows that Conjecture KS r is equivalent to PC if weassume equality in Equation 1.2 for all x ∈ l M 2 . Weaver further shows thatConjecture KS r is equivalent to PC even if we strengthen its assumptions soas to require that the vectors {ϕ i } M i=1 are of equal norm and that equalityholds in 1.2, but at great cost to our ɛ > 0.Conjecture 9 (KS ′ r ). There exists universal constants B ≥ 4 and ɛ > √ B sothat the following holds. Let {ϕ i } M i=1 be elements of lN 2 with ‖ϕ i ‖ = 1 fori = 1, 2, . . . , M and suppose for every x ∈ l N 2 ,M∑|〈x, ϕ i 〉| 2 = B‖x‖ 2 . (1.3)i=1Then, there is a partition {A j } r j=1 of {1, 2, . . . , M} so that for all x ∈ lM 2all j = 1, 2, . . . , r, ∑|〈x, ϕ i 〉| 2 ≤ (B − ɛ)‖x‖ 2 .i∈A jandWe introduce one more conjecture.Conjecture 10. There exist universal constants 0 < δ, √ δ ≤ ɛ < 1 and r ∈ Nso that for all N and all orthogonal projections P on l N 2 with δ(P ) ≤ δ