The Kadison-Singer and Paulsen Problems in Finite Frame Theory

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4 Peter G. Casazzaimplies PC follows by a minor modification of Propositions 7.6 and 7.7 of[1]. Then he introduces what he calls “Conjecture KS r ” (See Conjecture 8).A careful examination of the proof of Theorem 1 of [43] reveals that Weavershows Conjecture KS r implies Conjecture 1 which in turn implies KS which(after the theorem is proved) is equivalent to KS r .In [18] it was shown that PC fails for r = 2, even for projections withconstant diagonal 1/2. Recently [21] there appeared a frame theoretic concreteconstruction of non-2-pavable projections. If this construction can begeneralized, we would have a counterexample to PC and KS. We now lookat the construction from [21].Definition 2. A family of vectors {ϕ i } M i=1 for an N-dimensional Hilbertspace H N is (δ, r)-Rieszable if there is a partition {A j } r j=1 of {1, 2, . . . , M}so that for all j = 1, 2, . . . , r and all scalars {a i } i∈Aj we have‖ ∑i∈A ja i ϕ i ‖ 2 ≥ δ ∑i∈A j|a i | 2 .A projection P on H N is (δ, r)-Rieszable if {P e i } N i=1 is (δ, r)-Rieszable.We now have:Proposition 1. Let P be an orthogonal projection on H N . The following areequivalent:(1) The vectors {P e i } N i=1 are (δ, r)-Rieszable.(2) There is 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‖ ∑i∈A ja i (I − P )e i ‖ 2 ≤ (1 − δ) ∑(3) The matrix of I − P is (δ, r)-pavable.Proof. (1) ⇔ (2): For any scalars {a i } i∈AJ we have∑|a i | 2 = ‖ ∑a i P e i ‖ 2 + ‖ ∑i∈A j i∈A jHence,‖ ∑i∈A ja i (I−P )e i ‖ 2 ≤ (1−δ) ∑i∈A j|a i | 2 .i∈A ja i (I − P )e i ‖ 2 .i∈A j|a i | 2 if and only if ‖ ∑i∈A ja i P e i ‖ 2 ≥ δ ∑i∈A j|a i | 2 .(2) ⇔ (3): Given any partition {A j } r j=1 , any 1 ≤ j ≤ r and any x =∑i∈A ja i e i we have
1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 5〈(I − P )x, x〉 = ‖(I − P )x‖ 2 = ‖ ∑i∈A ja i (I − P )e i ‖ 2if and only if I − P ≤ (1 − δ)I.≤ (1 − δ) ∑i∈A j|a i | 2 = 〈(1 − δ)x, x〉,Given N ∈ N, let ω = exp ( 2πiN ) , we define the discrete Fourier transform(DFT) matrix in C N byD N =√1N(ωjk ) N−1j,k=0 .The main point of these D N matrices is that they √ are unitary matrices1for which the modulus of all the entries are equal toN. The following is asimple observation.Proposition 2. Let A = (a ij ) N i,j=1 be a matrix with orthogonal rows andsatisfying |a ij | 2 = a for all i, j. If we multiply the j th -row of A by a constantC j to get a new matrix B, then:(1) The rows of B are orthogonal.(2) The square sums of the entries of any column of B all equalaN∑Cj 2 .j=1(3) The square sum of the entries of the j th row of B equals aC 2 j .To construct our example, we start with a 2N × 2N DFT and multiplythe first N − 1 rows by √ √22 and the remaining rows byN+1to get a newmatrix B 1 . Next, we take a second 2N × 2N DFT √ matrix and multiply the2Nfirst N − 1 rows by 0 and the remaining rows by2N+1 to get a matrix B 2.We then put the matrices B 1 , B 2 side by side to get a N × 2N matrix B ofthe formB = (N-1) Rows √2 0√ √2(N+1) RowsN+12NN+1This matrix has 2N rows and 4N columns. Now we show that this matrixgives the required example.Proposition 3. The matrix B satisfies:(1) The columns are orthogonal and the square sum of the coefficients ofevery column equals 2.
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