The Kadison-Singer and Paulsen Problems in Finite Frame Theory

2 Peter G. CasazzaThe Kadison-Singer Problem and the Paulsen Problem. We warn the readerthat because we are restricting ourselves to finite dimensional Hilbert spaces,a significant body of literature on the infinite dimensional versions of theseproblems does not appear here.1.2 The Kadison-Singer ProblemFor over 50 years the Kadison-Singer Problem [36] has defied the best effortsof some of the most talented mathematicians of our time.Kadison-Singer Problem 1 (KS) Does every pure state on the (abelian)von Neumann algebra D of bounded diagonal operators on l 2 , the Hilbert spaceof square summable sequences on the integers, have a unique extension to a(pure) state on B(l 2 ), i.e., the von Neumann algebra of all bounded linearoperators on the Hilbert space l 2 ?A state of a von Neumann algebra R is a linear functional f on R for whichf(I) = 1 and f(T ) ≥ 0 whenever T ≥ 0 (whenever T is a positive operator).The set of states of R is a convex subset of the dual space of R which iscompact in the ω ∗ -topology. By the Krein-Milman theorem, this convex setis the closed convex hull of its extreme points. The extremal elements in thespace of states are called the pure states (of R).This problem arose from the very productive collaboration of Kadison andSinger in the 1950s when they were studying Dirac’s Quantum Mechanicsbook [30] which culminated in their seminal work on triangular operatoralgebras.It is now known that the 1959 Kadison-Singer Problem is equivalent tofundamental unsolved problems in a dozen areas of research in pure mathematics,applied mathematics and engineering (See [1, 2, 3, 4, 17, 26, 27, 37]and their references). We will not develop this topic in detail here since it isfundamentally an infinite dimensional problem and we are concentrating onfinite dimensional frame theory. In this chapter we will look at a number ofthese finite dimensional problems which are equivalent to KS and which areimpacted by finite frame theory. Most people today seem to agree with theoriginal statement of Kadison and Singer [36] that KS will have a negativeanswer and so all the equivalent forms will have negative answers also.1.2.1 The Paving ConjectureA significant advance on KS was made by Anderson [2] in 1979 when hereformulated KS into what is now known as the Paving Conjecture (Seealso [3, 4]). Lemma 5 of [36] shows a connection between KS and Paving. For

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 3notation, if J ⊂ {1, 2, . . . , n}, the diagonal projection Q J is the matrixwhose entries are all zero except for the (i, i) entries for i ∈ J which are allone. For a matrix A = (a ij ) N i,j=1 let δ(A) = max 1≤i≤N |a ii |.Definition 1. An operator T ∈ B(l N 2 ) is said to have an (r, ɛ)-paving if thereis a partition {A j } r j=1 of {1, 2, . . . , N} so that‖Q Aj T Q Aj ‖ ≤ ɛ‖T ‖.Paving Conjecture 1 (PC) For every 0 < ɛ < 1, there is a natural numberr so that for every natural number N and every linear operator T on l N 2whose matrix has zero diagonal, T has an (r, ɛ)-paving.It is important that r not depend on N in PC. We will say that an arbitraryoperator T satisfies PC if T − D(T ) satisfies PC where D(T ) is the diagonalof T .The only large classes of operators which have been shown to be pavableare “diagonally dominant” matrices [6, 7, 11, 31], l 1 -localized operators [24],matrices with all entries real and positive [32], matrices with small coefficientsin comparison with the dimension [13] (See [40] for a paving into blocks ofconstant size), and Toeplitz operators over Riemann integrable functions (Seealso [33]). Also, in [9] there is an analysis of the paving problem for certainSchatten C p -norms.Theorem 1. The Paving Conjecture has a positive solution if any one of thefollowing classes satisfies the Paving Conjecture:1. Unitary operators. [27]2. Orthogonal projections. [27]3. Orthogonal projections with constant diagonal 1/2. [18]4. Positive operators. [27]5. Self-adjoint operators. [27]6. Gram matrices (〈ϕ i , ϕ j 〉) i,j∈I where T : l 2 (I) → l 2 (I) is a bounded linearoperator, and T e i = ϕ i , ‖T e i ‖ = 1 for all i ∈ I. [27]7. Invertible operators (or invertible operators with zero diagonal). [27]8. Triangular operators [38]Recently, Weaver [43] provided important insight into KS by giving anequivalent problem to PC in terms of projections.Conjecture 1 (Weaver). There exist universal constants 0 < δ, ɛ < 1 andr ∈ N so that for all N and all orthogonal projections P on l N 2 with δ(P ) ≤ δ,there is a paving {A j } r j=1 of {1, 2, . . . , N} so that ‖Q A jP Q Aj ‖ ≤ 1 − ɛ, forall j = 1, 2, . . . , r.This needs some explanation since there is nothing in [43] that looks anythinglike Conjecture 1. Weaver observes that the fact that Conjecture 1

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.

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 .

8 Peter G. CasazzaSimilarly, ‖ ∑ i∈A ja i T e i ‖ 2 ≤ (1 + ɛ) ∑ i∈A j|a i | 2 .(2) ⇒ (3): This is obvious.(3) ⇒ (1): Let T ∈ B(l N 2 ) with T e i = ϕ i and ‖ϕ i ‖ = 1 for all 1 ≤ i ≤ N.By Theorem 1 part 6, it suffices to show that the Gram operator G of {ϕ i } N i=1is pavable. Fix 0 < δ < 1 and let ɛ > 0. Let ψ i = √ 1 − δ 2 ϕ i ⊕ δe i ∈ l N 2 ⊕ l N 2 .Then ‖ψ i ‖ = 1 for all 1 ≤ i ≤ N and for all scalars {a i } N i=1N∑N∑N∑δ |a i | 2 ≤ ‖ a i ψ i ‖ 2 = (1 − δ 2 )‖ a i T e i ‖ 2 + δ 2i=1i=1i=1≤ [ (1 − δ 2 )‖T ‖ 2 + δ 2] ∑N |a i | 2 .i=1N ∑i=1|a i | 2So {ψ i } N i=1 is a unit norm Riesz basic sequence and 〈ψ i, ψ k 〉 = (1−δ 2 )〈ϕ i , ϕ k 〉for all 1 ≤ i ≠ k ≤ N. By the R ɛ -Conjecture, there is a partition {A j } r j=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all x = ∑ i∈A ja i e i ,(1 − ɛ) ∑i∈A j|a i | 2 ≤ ‖ ∑a i ψ i ‖ 2 = 〈 ∑a i ψ i , ∑a k ψ k 〉i∈A j i∈A j k∈A j= ∑|a i | 2 ‖ψ i ‖ 2 +∑a i a k 〈ψ i , ψ k 〉i∈A j i≠k∈A j= ∑∑|a i | 2 + (1 − δ 2 ) a i a k 〈ϕ i , ϕ k 〉i∈A j i≠k∈A j= ∑|a i | 2 + (1 − δ 2 )〈Q Aj (G − D(G))Q Aj x, x〉i∈A j≤ (1 + ɛ) ∑|a i | 2 .i∈A jSubtracting ∑ i∈A j|a i | 2 through the inequality yields,That is,−ɛ ∑i∈A j|a i | 2 ≤ (1 − δ 2 )〈Q Aj (G − D(G))Q Aj x, x〉 ≤ ɛ ∑(1 − δ 2 )|〈Q Aj (G − D(G))Q Aj x, x〉| ≤ ɛ‖x‖ 2 .i∈A j|a i | 2 .Since Q Aj (G−D(G))Q Aj is a self-adjoint operator, we have (1−δ 2 )‖Q Aj (G−D(G))Q Aj ‖ ≤ ɛ, i.e., (1 − δ 2 )G (and hence G) is pavable.Remark 1. The proof of (3) ⇒ (1) of Theorem 2 illustrates a standard methodfor turning conjectures about unit norm Riesz basic sequences {ψ i } i∈I intoconjectures about unit norm familes {ϕ i } i∈I with T ∈ B(l 2 (I)) and T e i = ϕ i .Namely, given {ϕ i } i∈I and 0 < δ < 1 let ψ i = √ 1 − δ 2 f i ⊕δe i ∈ l 2 (I)⊕l 2 (I).

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 9Then {ψ i } i∈I is a unit norm Riesz basic sequence and for δ small enough, ψ iis close enough to ϕ i to pass inequalities from {ψ i } i∈I to {ϕ i } i∈I .The R ɛ -Conjecture is different from all other conjectures in this chapterin that it does not hold for equivalent norms on the Hilbert space in general.For example, if we renorm l 2 by: |{a i }| = ‖{a i }‖ l2 + sup i |a i | then the R ɛ -Conjecture fails for this equivalent norm. To see this, we proceed by way ofcontradiction and assume there is an 0 < ɛ < 1 and an r = r(ɛ, 2) satisfyingthe R ɛ -Conjecture. Let {e i } 2Ni=1 be the unit vectors for l2N2 and let x i =e 2i+e √2+1 2i+1for 1 ≤ i ≤ N. This is now a unit norm Riesz basic sequence withupper Riesz bound 2. Assume we partition {1, 2, . . . , 2N} into sets {A j } r j=1 .Then for some 1 ≤ k ≤ r we have |A k | ≥ N r . Let A ⊂ A k with |A| = N randa i = √ 1Nfor i ∈ A. Then| ∑ ( )1 √2 ra i x i | = √ + √i∈A 2 + 1 NSince the norm above is bounded away from one for large N, we cannot satisfythe requirements of the R ɛ -Conjecture. It follows that a positive solution toKS would imply a fundamental new result concerning “inner products”, notjust norms.Another important equivalent form of PC comes from [26]. This is, onits face, a significant weakening of the R ɛ -Conjecture while it still remainsequivalent to PC.Conjecture 4. There exists a constant A > 0 and a natural number r so thatfor all natural numbers N and all T : l N 2 → l N 2 with ‖T e i ‖ = 1 for alli = 1, 2, . . . , N and ‖T ‖ ≤ 2, there is a partition {A j } r j=1 of {1, 2, . . . , N} sothat for all j = 1, 2, . . . , r and all scalars {a i } i∈Aj we have‖ ∑i∈A ja i T e i ‖ 2 ≥ A ∑Theorem 3. Conjecture 4 is equivalent to PC.i∈A j|a i | 2 .Proof. Since PC is equivalent to the R ɛ -Conjecture, which in turn impliesConjecture 4, we just need to show that Conjecture 4 implies Conjecture1. So choose r, A satisfying Conjecture 4. Fix 0 < δ ≤ 3 4and let P be anorthogonal projection on l N 2 with δ(P ) ≤ δ Now, 〈P e i , e i 〉 = ‖P e i ‖ 2 ≤ δimplies ‖(I − P )e i ‖ 2 ≥ 1 − δ ≥ 1 4 . Define T : lN 2 → l N 2 by T e i =For any scalars {a i } N i=1 we have(I−P )ei‖(I−P )e i‖ .

10 Peter G. Casazza‖N∑a i T e i ‖ 2 N∑ a i=∥ ‖(I − P )ei=1i ‖ (I − P )e i∥N∑2≤a i∣‖(I − P )e i ‖ ∣i=1≤ 4i=1N∑|a i | 2 .i=1So ‖T e i ‖ = 1 and ‖T ‖ ≤ 2. By Conjecture 4, 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 T e i ‖ 2 ≥ A ∑i∈A j|a i | 2 .2Hence,∥ ∑∥∥∥∥∥2∥ ∑∥∥∥∥∥2a i (I − P )e i =a i ‖(I − P )e i ‖T e i∥i∈A j∥i∈A j≥ A ∑|a i | 2 ‖(I − P )e i ‖ 2i∈A j≥ A ∑|a i | 2 .4i∈A jIt follows that for all scalars {a i } i∈Aj ,∑|a i | 2 = ‖ ∑i∈A jNow, for all x = ∑ Ni=1 a ie i we havea i (I − P )e i ‖ 2i∈A j≥ ‖ ∑a i P e i ‖ 2 + A ∑|a i | 2 .4i∈A j i∈A j‖P Q Aj x‖ 2 = ‖ ∑i∈A ja i P e i ‖ 2 + ‖ ∑i∈A ja i P e i ‖ 2 ≤ (1 − A 4 ) ∑i∈A j|a i | 2 .Thus,‖Q Aj P Q Aj ‖ = ‖P Q Aj ‖ 2 ≤ 1 − A 4 .So Conjecture 1 holds.Weaver [43] established an important relationship between frames and PCby showing that the following conjecture is equivalent to PC.

12 Peter G. Casazzainto a finite number of subsets which were Riesz basic sequences. This led tothe conjecture:Feichtinger Conjecture 1 (FC) Every bounded frame (or equivalently,every unit norm frame) is a finite union of Riesz basic sequences.The finite dimensional form of FC looks like:Conjecture 7 (Finite Dimensional Feichtinger Conjecture). For every B, C >0, there is a natural number r = r(B, C) and a constant A = A(B, C) > 0so that whenever {ϕ i } N i=1 is a frame for HN with upper frame bound B and‖ϕ i ‖ ≥ C for all i = 1, 2, . . . , N, then {1, 2, . . . , N} can be partitioned intosubsets {A j } r j=1 so that for each 1 ≤ j ≤ r, {ϕ i} i∈Aj is a Riesz basic sequencewith lower Riesz basis bound A and upper Riesz basis bound B.There is a significant body of work on this conjecture [6, 7, 16, 31], yet, itremains open even for Gabor frames.We now check that the Feichtinger Conjecture is equivalent to PC.Theorem 5. The following are equivalent:(1) The Paving Conjecture.(2) The Feichtinger Conjecture.Proof. (1) ⇒ (2): Part (2) of Theorem 2 is equivalent to PC and clearlyimplies FC.(2) ⇒ (1): We will observe that FC implies Conjecture 4 which is equivalentto PC by Theorem 3. In Conjecture 4, {T e i } N i=1 is a frame for its spanwith upper frame bound 2. It is now immediate that the Finite DimensionalFeichtinger Conjecture implies Conjecture 4.Another equivalent formulation of KS due to Weaver [43].Conjecture 8 (KS r ). There are universal constants B and ɛ > 0 so that the followingholds. Let {ϕ i } M i=1 be elements of lN 2 with ‖ϕ i ‖ ≤ 1 for i = 1, 2, . . . , Mand suppose for every x ∈ l N 2 ,M∑|〈x, ϕ i 〉| 2 ≤ B‖x‖ 2 . (1.2)i=1Then, there is a partition {A j } r j=1 of {1, 2, . . . , M} so that for all x ∈ lN 2all j = 1, 2, . . . , r, ∑|〈x, ϕ i 〉| 2 ≤ (B − ɛ)‖x‖ 2 .i∈A jandTheorem 6. The following are equivalent:(1) The Paving Conjecture.(2) Conjecture KS r holds for some r ≥ 2.

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 13Proof. Assume Conjecture KS r is true for some fixed r, B, ɛ. We will showthat Conjecture 1 is true. Let P be an orthogonal projection on H M withδ(P ) ≤ 1 B. If P has rank N, then its range is a N-dimensional subspace W ofH M . Define ϕ i = √ B · P e i ∈ W for all 1 ≤ i ≤ M. We check‖ϕ i ‖ 2 = B · ‖P e i ‖ 2 = B〈P e i , e i 〉 ≤ Bδ(P ) ≤ 1, for all i = 1, 2, . . . , M.Now, if x ∈ W is any unit vector thenM∑M∑|〈x, ϕ i 〉| 2 = |〈x, √ M∑BP e i 〉| 2 = B · |〈x, e i 〉| 2 = B.i=1i=1By Conjecture KS r , there is a partition {A j } r j=1 of {1, 2, . . . , M} satisfyingfor all 1 ≤ j ≤ r and all unit vectors x ∈ W ,∑|〈x, ϕ i 〉| 2 ≤ B − ɛ.i∈A jThen ∑ rj=1 Q A j= Id, and for any unit vector x ∈ W we havei=1i=1M∑M∑‖Q Aj P x‖ 2 = |〈Q Aj P x, e i 〉| 2 = |〈x, P Q j e i 〉| 2= 1 ∑|〈x, ϕ i 〉| 2 ≤ ɛ BB .i∈A jThus Conjecture 1 holds and so PC holds.Conversely, assume KS r fails for all r. Fix B = r ≥ 2 and let {ϕ i } M i=1in H N be a counterexample with ɛ = 1. Let ψ i = √ ϕiBand note that‖ψ i ψ T i ‖ = ‖ψ i ‖ 2 ≤ 1 B , for all i = 1, 2, . . . , M and ∑ Mi=1 ψ iψ T i ≤ Id. ThenId − ∑ Mi=1 ψ iψi T is a positive finite rank operator, so we can find positiverank one operators ψ i ψiT for M + 1 ≤ i ≤ K such that ‖ψ i ψi T ‖ ≤ 1 Bfor all1 ≤ i ≤ K and ∑ Ki=1 ψ iψi T = Id.Let T be the analysis operator for {ψ i } K i=1 , which is an isometry and if Pis the orthogonal projection of H K with range T (H N ), then P e i = T ψ i forall i = 1, 2, . . . , K. Let D be the diagonal matrix with the same diagonal asP . Then‖D‖ = max ‖ψ i‖ 2 ≤ 11≤i≤K B .Let {Q j } r j=1 be any K ×K diagonal projections which sum to the identity.Define a partition {A j } r j=1 of {1, 2, . . . , K by letting A j be the diagonal of Q j .By our choice of {ϕ i } M i=1 , there exists 1 ≤ j ≤ r and x ∈ HN with ‖x‖ = 1andi=1

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

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 15and ‖P e i ‖ = ‖P e j ‖ for all i, j = 1, 2, . . . , N, there is a paving {A j } r j=1 of{1, 2, . . . , N} so that ‖Q Aj P Q Aj ‖ ≤ 1 − ɛ, for all j = 1, 2, . . . , r.Using Conjecture 9 we can see that PC is equivalent to Conjecture 10.Theorem 7. PC is equivalent to Conjecture 10.Proof: It is clear that Conjecture 1 (which is equivalent to (PC)) impliesConjecture 10. So we assume that Conjecture 10 holds and we will showthat Conjecture 9 holds. Let {ϕ i } M i=1 be elements of HN with ‖ϕ i ‖ = 1 fori = 1, 2, . . . , M and suppose for every x ∈ H N ,M∑|〈x, ϕ i 〉| 2 = B‖x‖ 2 , (1.4)i=1where 1 1B≤ δ. It follows from Equation 1.4 that { √Bϕ i } M i=1 is an equal normParseval frame and so by Naimark’s Theorem, we may assume there is alarger Hilbert space l M 2 and a projection P : l M 2 → H N so that P e i = ϕ i forall i = 1, 2, . . . , M. Now ‖P e i ‖ 2 = 〈P e i , e i 〉 = 1 B≤ δ for all i = 1, 2, . . . , N.So by Conjecture 10, there is a paving {A j } r j=1 of {1, 2, . . . , M} so that‖Q Aj P Q Aj ‖ ≤ 1 − ɛ, for all j = 1, 2, . . . , r. Now for all 1 ≤ j ≤ r and allx ∈ l N 2 we have:It follows that for all x ∈ l N 2∑‖Q Aj P x‖ 2 ==M∑|〈Q Aj P x, e i 〉| 2i=1M∑|〈x, P Q Aj e i 〉| 2i=1= 1 ∑|〈x, ϕ i 〉| 2Bi∈A jwe have≤ ‖Q Aj P ‖ 2 ‖x‖ 2= ‖Q Aj P Q Aj ‖‖x‖ 2≤ (1 − ɛ)‖x‖ 2 .i∈A j|〈xϕ i 〉| 2 ≤ (B − ɛB)‖x‖ 2 .Since ɛB > √ B, we have verified Conjecture 9.⊓⊔

16 Peter G. Casazza1.2.4 The Bourgain-Tzafriri ConjectureWe start with a fundamental theorem of Bourgain and Tzafriri called theRestricted Invertibility Principle. This theorem led to the (strong and weak)Bourgain-Tzafriri Conjectures. We will see that these conjectures are equivalentto PC.In 1987, Bourgain and Tzafriri [12] proved a fundamental result in Banachspace theory known as the Restricted Invertibility Principle.Theorem 8 (Bourgain-Tzafriri). There is a universal constants 0 < c < 1so that whenever T : l N 2 → l N 2 is a linear operator for which ‖T e i ‖ = 1,for 1 ≤ i ≤ N, then there exists a subset σ ⊂ {1, 2, . . . , N} of cardinality|σ| ≥ cN/‖T ‖ 2 so that for all choices of scalars {a j } j∈σ ,‖ ∑ a j T e j ‖ 2 ≥ c ∑j∈σj∈σ|a j | 2 .A close examination of the proof of the theorem [12] yields that c is onthe order of 10 −72 . The proof of the theorem uses probabilistic and functionanalytic techniques, is non-trivial and is non-constructive. A significant breakthroughoccurred recently when Spielman and Srivastava [39] presented analgorithm for proving the Restricted-Invertibility Theorem. Moreover, theirproof gives the best possible constants in the theorem.Theorem 9 (Restricted Invertibility Theorem: Spielman-SrivastavaForm). Assume {v i } M i=1 are vectors in lN 2 with A = ∑ Mi=1 v ivi T = I and0 < ɛ < 1. If L : l N 2 → l N 2 is a linear operator, then there is a subset J ⊂{1, 2, . . . , M} of size |J| ≥ ɛ 2 ‖L‖ 2 F‖L‖ 2and( )∑λ min Lv i (Lv i ) Ti∈Jfor which {Lv i } i∈J is linearly independent> (1 − ɛ)2 ‖L‖ FMwhere ‖L‖ F is the Frobenious norm of L and λ min is the smallest eigenvalueof the operator computed on span {v i } i∈J .This generalized form of the Restricted-Invertibility Theorem was introducedby Vershynin [41] where he studied the contact points of convex bodiesusing John’s decompositions of the identity. The corresponding theorem forinfinite dimensional Hilbert spaces is still open. But this case requires the setJ to be large with respect to the Beurling density [6, 7]. Special cases of thisproblem were solved in [24, 41].The inequality in the Restricted Invertibility Theorem is referred to as alower l 2 -bound. It is known [28, 41] that there is a corresponding close to oneupper l 2 -bound which can be achieved in the theorem. Recently, a two-sidedSpielman-Srivastava algorithm was given [25] which gives the best two sidedlower and upper bounds in BT.,

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 17Corollary 1. For T : l N 2 → l N 2 and an orthonormal basis {e i } N i=1 with‖T e i ‖ 2 = 1, i = 1, . . . , N, and for any α ∈ (0, 1) there exists a subsetJ α ⊆ {1, 2, . . . , N}, satisfying⌊ ⌋ α2N1. |J α | ≥2 ‖T ‖ 2 ,and2. The condition number of T | span {ej: j∈J α} is bounded above by (1 − α) −4 .The corresponding theorem for infinite dimensional Hilbert spaces is stillopen. Special cases of this problem were solved in [24, 41].Theorem 8 gave rise to a problem in the area which has received a greatdeal of attention [13, 26, 27].Bourgain-Tzafriri Conjecture 1 (BT) There is a universal constant A >0 so that for every B > 1 there is a natural number r = r(B) satisfying: Forany natural number N, if T : l N 2 → l N 2 is a linear operator with ‖T ‖ ≤ Band ‖T e i ‖ = 1 for all i = 1, 2, . . . , N, then there is a partition {A j } r j=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all choices of scalars {a i } i∈Ajwe have:‖ ∑a i T e i ‖ 2 ≥ A ∑|a i | 2 .i∈A j i∈A jSometimes BT is called strong BT since there is a weakening called weakBT. In weak BT we allow A to depend upon the norm of the operator T . Asignificant amount of effort over the years was invested in trying to show thatstrong and weak BT are equivalent. Casazza and Tremain finally proved thisequivalence [26]. We will not have to do any work here since we developed allof the needed results in earlier sections.Theorem 10. The following are equivalent:(1) The Paving Conjecture.(2) The Bourgain-Tzafriri Conjecture.(3) The (weak) Bourgain-Tzafriri Conjecture.Proof. (1) ⇒ (2) ⇒ (3): The Paving Conjecture is equivalent to the R ɛ -Conjecture which clearly implies the Bourgain-Tzafriri Conjecture and thisimmediately implies the (weak) Bourgain-Tzafriri Conjecture.(3) ⇒ (1): The weak Bourgain-Tzafriri Conjecture immediately impliesConjecture 4 which is equivalent to the Paving Conjecture.1.2.5 Partitioning Frames into Frame SubsetsA natural and frequently occurring problem in frame theory is to partitiona frame into subsets each of which has good frame bounds. This innocent

18 Peter G. Casazzalooking question turns out to be much deeper than it looks, and as we willnow see, it is equivalent to PC.Conjecture 11. There exists an ɛ > 0 so that for large K, for all N and allequal norm Parseval frames {ϕ i } KNi=1 for lN 2 , there is a J ⊂ {1, 2, . . . , KN} sothat both {ϕ i } i∈J and {ϕ i } i∈J c have lower frame bounds which are greaterthan ɛ.The ideal situation would be for Conjecture 11 to hold for all K ≥ 2. Inorder for {ϕ i } i∈J and {ϕ i } i∈J c to both be frames for l N 2 , they at least haveto span l N 2 . So the first question is whether we can partition our frame intospanning sets. This follows from a generalization of the Rado-Horn theorem.See the chapter on independence and spanning properties of frames.Proposition 4. Every equal norm Parseval frame {ϕ i } KN+Li=1 , 0 ≤ L < Nfor l N 2 can be partitioned into K linearly independent spanning sets plus alinearly independent set of L elements.The natural question is whether we can do such a partition so that eachof the subsets has good frame bounds, i.e., a universal lower frame bound forall subsets. Before addressing this question, we state another conjecture.Conjecture 12. There exists ɛ > 0 and a natural number r so that for allN, all large K and all equal norm Parseval frames {ϕ i } KNi=1 in lN 2 there is apartition {A j } r j=1 of {1, 2, . . . , KN} so that for all j = 1, 2, . . . , r the Besselbound of {ϕ i } i∈Aj is ≤ 1 − ɛ.We will now establish a relationship between our conjectures and PC.Theorem 11. (1) Conjecture 11 implies Conjecture 12.(2) Conjecture 12 is equivalent to PC.Proof. (1): Fix ɛ > 0, r, K as in Conjecture 11. Let {ϕ i } KNi=1 be an equal normParseval frame for an N-dimensional Hilbert space H N . By Naimark’s Theoremwe may assume there is an orthogonal projection P on l KN2 with P e i = ϕ ifor all i = 1, 2, . . . , KN. By Conjecture 11, there is a J ⊂ {1, 2, . . . , KN} sothat {P e i } i∈J and {P e i } i∈J c both have a lower frame bound of ɛ > 0. Hence,for x ∈ H M = P (l KN2 ),∑KN‖x‖ 2 = |〈x, P e i 〉| 2 = ∑ |〈x, P e i 〉| 2 + ∑ |〈x, P e i 〉| 2i=1i∈Ji∈J c≥ ∑ i∈J|〈x, P e i 〉| 2 + ɛ‖x‖ 2 .That is, ∑ i∈J |〈x, P e i〉| 2 ≤ (1−ɛ)‖x‖ 2 . So the upper frame bound of {P e i } i∈J(which is the norm of the analysis operator (P Q J ) ∗ for this frame) is ≤ 1 − ɛ.Since P Q J is the synthesis operator for this frame, we have that ‖Q J P Q J ‖ =

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 19‖P Q J ‖ 2 = ‖(P Q J ) ∗ ‖ 2 ≤ 1 − ɛ. Similarly, ‖Q J cP Q J c‖ ≤ 1 − ɛ. So Conjecture12 holds for r = 2.(2): We will show that Conjecture 12 implies Conjecture 5. Choose ɛ andr satisfying Conjecture 12 for all large K. In particular, choose any K with1 √K< α. Let {ϕ i } M i=1 be a unit norm K-tight frame for an N-dimensionalHilbert space H N . Then M = ∑ Mi=1 ‖ϕ i‖ 2 = KN. Since { √ 1Kϕ i } M i=1 is anequal norm Parseval frame, by Naimark’s Theorem we may assume there isan orthogonal projection P on l M 2 with P e i = √ 1Kϕ i , for i = 1, 2, . . . , M. ByConjecture 12 there is a partition {A j } r j=1 of {1, 2, . . . , M} so that the Besselbound ‖(P Q Aj ) ∗ ‖ 2 for each family {ϕ i } i∈Aj is ≤ 1 − ɛ. So for j = 1, 2, . . . , rand any x ∈ l N 2 we have∑i∈A j|〈x,1√ ϕ i 〉| 2 = ∑Ki∈A j|〈x, P Q Aj e i 〉| 2 = ∑i∈A j|〈Q Aj P x, e i 〉| 2 ≤ ‖Q Aj P x‖ 2≤ ‖Q Aj P ‖ 2 ‖x‖ 2 = ‖(P Q Aj ) ∗ ‖ 2 ‖x‖ 2 ≤ (1 − ɛ)‖x‖ 2 .Hence,∑i∈A j|〈x, ϕ i 〉| 2 ≤ K(1 − ɛ)‖x‖ 2 = (K − Kɛ)‖x‖ 2 .Since Kɛ > √ K, we have verified Conjecture 5.For the converse, choose r, δ, ɛ satisfying Conjecture 1. If {ϕ i } KNi=1 is anequal norm Parseval frame for an N-dimensional Hilbert space H N with 1 K ≤δ, by Naimark’s Theorem we may assume we have an orthogonal projectionP on l KN2 with P e i = ϕ i for i = 1, 2, . . . , KN. Since δ(P ) = ‖ϕ i ‖ 2 ≤ 1 K ≤ δ,by Conjecture 1 there is a partition {A j } r j=1 of {1, 2, . . . , KN} so that for allj = 1, 2, . . . , r,‖Q Aj P Q Aj ‖ = ‖P Q Aj ‖ 2 = ‖(P Q Aj ) ∗ ‖ 2 ≤ 1 − ɛ.Since ‖(P Q Aj ) ∗ ‖ 2 is the Bessel bound for {P e i } i∈Ajthat Conjecture 12 holds.= {ϕ i } i∈Aj , we have1.3 The Sundberg ProblemRecently, an apparent weakening of the Kadison-Singer Problem has arisen.In his work on interpolation in complex function theory, Sundberg noticedthe following problem. Although this is an infinite dimensional problem, westate it here because of its connections to the Kadison-Singer Problem.Problem 1 (Sundberg Problem). If {ϕ} ∞ i=1 is a unit norm Bessel sequence,can we partition {ϕ i } ∞ i=1 into a finite number of spanning sets?

20 Peter G. CasazzaThis problem appears to be quite innocent, but it is surprisingly difficult. Itis immediate that the Feichtinger Conjecture implies the Sundberg Problem.Theorem 12. A positive solution to the Feichtinger Conjecture implies apositive solution to the Sundberg Problem.Proof. If {ϕ i } ∞ i=1 is a unit norm Bessel sequence then by FC, we can partitionthe natural numbers into a finite number of sets {A j } r j=1 so that {ϕ i} i∈Aj isa Riesz sequence for all j = 1, 2, . . . , r. For each j = 1, 2, . . . , r choose i j ∈ A j .Then neither ϕ ij nor {ϕ i } i∈Aj\{i j} can span the space.1.4 The Paulsen ProblemThe Paulsen Problem has been intractable for over a dozen years despitereceiving quite a bit of attention. In this section we look at the current stateof the art on this problem. First we need two definitions.Definition 4. A frame {ϕ i } M i=1 for HN with frame operator S is said to beɛ-nearly equal norm if(1 − ɛ) N M ≤ ‖ϕ i‖ 2 ≤ (1 + ɛ) N , for all i = 1, 2, . . . , M,Mand it is ɛ-nearly Parseval ifWe also need:(1 − ɛ)Id ≤ S ≤ (1 + ɛ)Id.Definition 5. Given frames Φ = {ϕ i } M i=1 and Ψ = {ψ i} M i=1 for HN , we definethe distance between them byd(Φ, Ψ) =M∑‖ϕ i − ψ i ‖ 2 .i=1The above is not exactly a distance function since we have not taken thesquare root of the right-hand-side of the equality. But since this formulationis standard, we will use it. We can now state the Paulsen Problem.Problem 2 (Paulsen Problem). How close is an ɛ-nearly equal norm andɛ-nearly Parseval frame to an equal norm Parseval frame?The importance of the Paulsen Problem is that we have algorithms forconstructing frames which are equal norm and nearly Parseval. The questionis, if we work with these frames, are we sure that we are working with a framewhich is close to some equal norm Parseval frame? We are looking for the

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 21function f(ɛ, N, M) so that every ɛ-nearly equal norm and ɛ-nearly Parsevalframe Φ = {ϕ i } M i=1 satisfies d(Φ, Ψ) ≤ f(ɛ, N, M),for some equal norm Parseval frame Ψ. A simple compactness argument dueto Hadwin (See [10]) shows that such a function must exist.Lemma 1. The function f(ɛ, N, M) exists.Proof. We will proceed by way of contradiction. If this fails, then there is an0 < ɛ so that for every δ = 1 n , there is a frame {ϕn i }M i=1 with frame bounds1 − 1 n , 1 + 1 nand satisfying(1 −n) 1 (NM ≤ ‖ϕn i ‖ ≤ 1 +n) 1 NM ,while Φ n = {ϕ n i }M i=1 is a distance greater than ɛ from any equal norm Parsevalframe. By compactness and switching to a subsequence, we may assume thatlimn→∞ ϕn i = ϕ i , exists for all i = 1, 2, . . . , M.But now Φ = {ϕ i } M i=1 is an equal norm Parseval frame contradicting the factthatd(Φ n , Φ) ≥ ɛ > 0, for all n = 1, 2, . . . ,for any equal norm Parseval frame.The problem with this argument is that it does not give any quantitativeestimate on the parameters. We do not have a good idea of what form thefunction f(ɛ, N, M) must have. We do not even know if M needs to be inthe function or if it is independent of the number of frame vectors. Thefollowing example shows that the Paulsen function is certainly a function ofthe dimension of the space.Lemma 2. The Paulsen function satisfiesf(ɛ, N, M) ≥ ɛ 2 N.Proof. Fix an ɛ > 0 and an orthonormal basis {e j } N j=1 for HN . We define aframe {ϕ i } 2Ni=1 by ϕ i ={ 1−ɛ √2e i1+ɛ √2e i−Nif 1 ≤ i ≤ Nif N + 1 ≤ i ≤ 2NBy the definition, {ϕ i } M i=1 is ɛ-nearly equal norm. Also, for any x ∈ HN wehave

22 Peter G. Casazza2N∑i=1|〈x, ϕ i 〉| 2 =(1 − ɛ)22N∑|〈x, e i 〉| 2 +i=1(1 + ɛ)22N∑|〈x, e i 〉| 2 = (1 + ɛ 2 )‖x‖ 2 .i=1So {ϕ i } 2Ni=1 is a (1 + ɛ2 ) tight frame and hence an ɛ-nearly Parseval frame.The closest equal norm frame to {ϕ i } 2N eii=1 is { √2} N i=1 ∪ { √ ei2} N i=1 . Also,N∑i=1‖ e i√ −ϕ i ‖ 2 + 22N∑i=N+1‖ e i−N√ −ϕ i ‖ 2 = 2N∑i=1‖ ɛ √2e i ‖ 2 +‖N∑i=1‖ ɛ √2e i ‖ 2 = ɛ 2 N.The main difficulty with solving the Paulsen Problem is that finding aclose equal-norm frame to a given frame involves finding a close frame whichsatisfies a geometric condition while finding a close Parseval frame to a givenframe involves satisfying a (algebraic) spectral condition. At this time, welack techniques for combining these two conditions. However, each of themindividually has a known solution. That is, we do know the closest equalnorm frame to a given frame [15] and we do know the closest Parseval frameto a given frame [5, 10, 15, 22, 35].Lemma 3. If {ϕ i } M i=1 is an ɛ-nearly equal norm frame in HN , then the closestequal norm frame to {ϕ i } M i=1 iswhereψ i = a ϕ i, for i = 1, 2, . . . , M,‖ϕ i ‖∑ Mi=1a =‖ϕ i‖.MIt is well known that for a frame {ϕ i } M i=1 for HN with frame operator S,the closest Parseval frame to {ϕ i } M i=1 is {S−1/2 ϕ i } M i=1 [5, 10, 15, 22, 35]. Wewill give the version from [10] here.Proposition 5. Let {ϕ i } M i=1 be a frame for a N-dimensional Hilbert spaceH N , with frame operator S = T ∗ T . Then {S −1/2 ϕ i } M i=1 is the closest Parsevalframe to {ϕ i } M i=1 . Moreover, if {ϕ i} M i=1 is an ɛ-nearly Parseval frame thenM∑‖S −1/2 ϕ i − ϕ i ‖ 2 ≤ N(2 − ɛ − 2 √ 1 − ɛ) ≤ Nɛ 2 /4.i=1Proof. We first check that {S −1/2 ϕ i } M i=1 is the closest Parseval frame to{ϕ i } M i=1 .The squared l 2 -distance between {ϕ i } M i=1 and any Parseval frame {ψ j} n j=1can be expressed in terms of their analysis operators T and T 1 as‖F − G‖ 2 = T r[(T − T 1 )(T − T 1 ) ∗ ]= T r[T T ∗ ] + T r[T 1 T ∗ 1 ] − 2RT r[T T ∗ 1 ]

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 23Choosing a Parseval frame {ψ i } M i=1 is equivalent to choosing the isometry T 1.To minimize the distance over all choices of T 1 , consider the polar decompositionT = UP , where P is positive and U is an isometry. In fact, S = T ∗ Timplies P = S 1/2 , which means its eigenvalues are bounded away from zero.Since P is positive and bounded away from zero, the term T r[T T ∗ 1 ] =T r[UP T ∗ 1 ] = T r[T ∗ 1 UP ] is an inner product between T 1 and U. Its magnitudeis bounded by the Cauchy Schwarz inequality, and thus it has a maximal realpart if T 1 = U which implies T ∗ 1 U = I. In this case, T = T 1 P = T 1 S 1/2 ,or equivalently T ∗ 1 = S −1/2 T ∗ , and we conclude ψ i = S −1/2 ϕ i for all i =1, 2, . . . M.After choosing T 1 = T S −1/2 , the l 2 -distance is expressed in terms of theeigenvalues {λ j } N j=1 of S = T ∗ T by‖F − G‖ 2 = T r[S] + T r[I] − 2T r[S 1/2 ]N∑N∑ √= λ j + N − 2 λj .j=1If 1 − ɛ ≤ λ j ≤ 1 + ɛ for all j = 1, 2, . . . N, calculus shows that the maximumof λ j − 2 √ λ j is achieved when λ j = 1 − ɛ.Consequently,j=1‖F − G‖ 2 ≤ 2N − Nɛ − 2N √ 1 − ɛ.Estimating √ 1 − ɛ by the first three terms in its decreasing power series givesthe inequality ‖F − G‖ 2 ≤ Nɛ 2 /4.It can be shown that the estimate above is exact and so we have separateverification that the closeness function is a function of N.There is a simple algorithm for turning any frame into an equal normframe with the same frame operator due to Holmes and Paulsen [34].Proposition 6. There is an algorithm for turning any frame into an equalnorm frame with the same frame operator.Proof. Let {ϕ i } M i=1 be a frame for HN with frame operator S and analysisoperator T . ThenM∑‖ϕ i ‖ 2 = T r S.i=1Let λ = T r SM . If ‖ϕ i‖ 2 = λ, for all m = 1, 2, . . . , M, then we are done.Otherwise, there exists 1 ≤ i ≠ j ≤ M with ‖ϕ i ‖ 2 > λ > ‖ϕ j ‖ 2 . For any θ,replace the vectors ϕ i , ϕ j by the vectorsψ i = (cosθ)ϕ i − (sinθ)ϕ j , ψ j = (sinθ)ϕ i + (cosθ)ϕ j , ψ k = ϕ k , for k ≠ i, j.Now, the analysis operator for {ψ i } M i=1 is T 1 = UT for a unitary operator Uon l N 2 given by the Givens’ rotation. Hence, T1 ∗ T 1 = T ∗ U ∗ UT = T ∗ T = S; so

24 Peter G. Casazzathe frame operator is unchanged for any value of θ. Now choose the θ yielding‖ψ i ‖ 2 = λ. Repeating this process at most M − 1 times yields an equal normframe with the same frame operator as {ϕ i } M i=1 .Using a Parseval frame in Proposition 6, we do get to an equal normParseval frame. The problem, again, is that we do not have any quantitativemeasure of how close these two Parseval frames are.There is an obvious approach towards solving the Paulsen Problem. Givenan ɛ-nearly equal norm ɛ-nearly Parseval frame {ϕ i } M i=1 for HN with frameoperator S, we can switch to the closest Parseval frame {S −1/2 ϕ i } M i=1 . Thenswitch to the closest equal norm frame to {S −1/2 ϕ i } M i=1 , call it {ψ i} M i=1 withframe operator S 1 . Now switch to {S −1/21 ψ i } M i=1 and again switch to the closestequal norm frame and continue. Unfortunately, even if we could show thatthis process converges and we could check the distance traveled through thisprocess, we would still not have an answer to the Paulsen Problem becausethis process does not have to converge to an equal norm Parseval frame. Inparticular, there is a fixed point of this process which is not an equal normParseval frame.Example 1. Let {e i } N i=1 be an orthonormal basis for lN 2 and let {ϕ i } N+1i=1 bea equiangular unit norm tight frame for l N 2 . Then {e i ⊕ 0} N i=1 ∪ {0 ⊕ ϕ i} N+1i=1in l N 2 ⊕ l N 2 is a ɛ = 1 N -nearly equal norm and 1 N- nearly Parseval frame withframe operator, say S. A direct calculation shows that taking S −1/2 of theframe vectors and switching to the closest equal norm frame leaves the frameunchanged.The Paulsen Problem has proven to be intractable for over 12 years. Recently,two partial solutions to the problem were given in [10, 20] each havingsome advantages. Since each of these papers is technical, we will only outlinethe ideas here.In [10], a new technique is introduced. This is a system of vector valuedODE’s which starts with a given Parseval frame and has the property that allframes in the flow are still Parseval while approaching an equal norm Parsevalframe. They then bound the arc length of the system of ODE’s by the frameenergy. Finally, giving an exponential bound on the frame energy, they havea quantitative estimate for the distance between the initial, ɛ-nearly equalnormand ɛ-nearly Parseval frame F and the equal-norm Parseval frame G.For the method to work, they must assume that the dimension N of theHilbert space and the number M of frame vectors are relatively prime. Theauthors show that in practice, this is not a serious restriction. The main resultof [10] isTheorem 13. Let N, M ∈ N be relatively prime, let 0 < ɛ < 1 2, and assumeΦ = {ϕ i } M i=1 is an ɛ-nearly equal-norm and ɛ-nearly Parseval frame for areal or complex Hilbert space of dimension N. Then there is an equal-normParseval frame Ψ = {ψ i } M i=1 such that

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 25‖Φ − Ψ‖ ≤ 298 N 2 M(M − 1) 8 ɛ.In [20], the authors present a new iterative algorithm—gradient descent ofthe frame potential—for increasing the degree of tightness of any finite unitnorm frame. The algorithm itself is trivial to implement, and it preservescertain group structures present in the initial frame. In the special case wherethe number of frame elements is relatively prime to the dimension of theunderlying space, they show that this algorithm converges to a unit normtight frame at a linear rate, provided the initial unit norm frame is alreadysufficiently close to being tight. The main difference between this approachand the approach in [10] is that in [10], the authors start with a nearly equalnorm Parseval frame and improve its closeness to an equal norm frame whilemaintaining Parseval, and in [20] the authors start with an equal norm nearlyParseval frame and give an algorithm for improving its algebraic propertieswhile changing its transform as little as possible. The main result from [20]is:1Theorem 14. Suppose M and N are relatively prime. Pick t ∈ (0,2M), andlet Φ 0 = {ϕ i } M i=1 be a unit norm frame with analysis operator T 0 satisfying‖T0 ∗ T 0 − M N I‖2 HS ≤ 2N. Now, iterate the gradient descent of the frame potential3method to obtain Φ k . Then, Φ ∞ := lim k Φ k exists and is a unit norm tightframe satisfying‖Φ ∞ − Φ 0 ‖ HS ≤ 4N 20 M 8.51−2Mt∥ T∗0 T 0 − M N I∥ ∥HS.In [10], the authors showed there is a connection between the PaulsenProblem and a fundamental open problem in operator theory.Problem 3 (Projection Problem). Let H N be an N-dimensional Hilbertspace with orthonormal basis {e i } N i=1 . Find the function g(ɛ, N, M) satisfyingthe following. If P is a projection of rank M on H N satisfying(1 − ɛ) M N ≤ ‖P e i‖ 2 ≤ (1 + ɛ) M , for all i = 1, 2, . . . , N,Nthen there is a projection Q with ‖Qe i ‖ 2 = M Nfor all i = 1, 2, . . . , N satisfyingN∑‖P e i − Qe i ‖ 2 ≤ g(ɛ, N, M).i=1In [14], it was shown that the Paulsen problem is equivalent to the ProjectionProblem and that their closeness functions are within a factor of 2 ofone another. The proof of this result gives several exact connections betweenthe distance between frames and the distance between the ranges of theiranalysis operators.

26 Peter G. CasazzaTheorem 15. Let Φ = {ϕ i } i∈I , Ψ = {ψ i } i∈I be Parseval frames for a Hilbertspace H with analysis operators T 1 , T 2 respectively. Ifd(Φ, Ψ) = ∑ i∈I‖ϕ i − ψ i ‖ 2 < ɛ,thend(T 1 (Φ), T 2 (Ψ)) = ∑ i∈I‖T 1 ϕ i − T 2 ψ i ‖ 2 < 4ɛ.Proof. Note that for all j ∈ I,T 1 ϕ j = ∑ i∈I〈ϕ j , ϕ i 〉e i , and T 2 ψ j = ∑ i∈I〈ψ j , ψ i 〉e i .Hence,‖T 1 ϕ j − T 2 ψ j ‖ 2 = ∑ |〈ϕ j , ϕ i 〉 − 〈ψ j , ψ i 〉| 2i∈I= ∑ |〈ϕ j , ϕ i − ψ i 〉 + 〈ϕ j − ψ j , ψ i 〉| 2i∈I≤ 2 ∑ |〈ϕ j , ϕ i − ψ i 〉| 2 + 2 ∑ |〈ϕ j − ψ j , ψ i 〉| 2 .i∈Ii∈ISumming over j and using the fact that our frames Φ and Ψ are Parsevalgives∑‖T 1 ϕ j − T 2 ψ j ‖ 2 ≤ 2 ∑ ∑|〈ϕ j , ϕ i − ψ i 〉| 2 + 2 ∑ ∑|〈ϕ j − ψ j , ψ i 〉| 2j∈Ij∈I i∈Ij∈I i∈I= 2 ∑ ∑|〈ϕ j , ϕ i − ψ i 〉| 2 + 2 ∑ ‖ϕ j − ψ j ‖ 2i∈I j∈Ij∈I= 2 ∑ ‖ϕ i − ψ i ‖ 2 + 2 ∑ ‖ϕ j − ψ j ‖ 2i∈Ij∈I= 4 ∑ ‖ϕ j − ψ j ‖ 2 .j∈INext, we want to relate the chordal distance between two subspaces to thedistance between their orthogonal projections. First we need to define thedistance between projections.Definition 6. If P, Q are projections on H N , we define the distance betweenthem byM∑d(P, Q) = ‖P e i − Qe i ‖ 2 ,i=1where {e i } N i=1 is an orthonormal basis for HN .

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 27The chordal distance between subspaces of a Hilbert space was defined by[29] and has been shown to have many uses over the years.Definition 7. Given M-dimensional subspaces W 1 , W 2 of a Hilbert space,define the M-tuple (σ 1 , σ 2 , . . . , σ M ) as follows:σ 1 = max{〈x, y〉 : x ∈ Sp W1, y ∈ Sp W2} = 〈x 1 , y 1 〉,where Sp W is the unit sphere of the subspace W . For 2 ≤ i ≤ M,σ i = max{〈x, y〉 : ‖x‖ = ‖y‖ = 1, 〈x j , x〉 = 0 = 〈y j , y〉, for 1 ≤ j ≤ i − 1},whereσ i = 〈x i , y i 〉.The M-tuple (θ 1 , θ 2 , . . . , θ M ) with θ i = cos −1 (σ i ) is called the principleangles between W 1 , W 2 . The chordal distance between W 1 , W 2 is given byd 2 c(W 1 , W 2 ) =M∑sin 2 θ i .i=1So by the definition, there exists orthonormal bases {a j } M j=1 , {b j} M j=1W 1 , W 2 respectively satisfying( θ‖a j − b j ‖ = 2sin , for all j = 1, 2, . . . , M.2)forIt follows that for 0 ≤ θ ≤ π 2 ,Hence,( ) θsin 2 θ ≤ 4sin 2 = ‖a j − b j ‖ 2 ≤ 4sin 2 θ, for all j = 1, 2, . . . , M.2d 2 c(W 1 , W 2 ) ≤M∑‖a j − b j ‖ 2 ≤ 4d 2 c(W 1 , W 2 ). (1.5)j=1We also need the following result [29].Lemma 4. If H N is an N-dimensional Hilbert space and P, Q are rank Morthogonal projections onto subspaces W 1 , W 2 respectively, then the chordaldistance d c (W 1 , W 2 ) between the subspaces satisfiesd 2 c(W 1 , W 2 ) = M − T r P Q.Next we give a precise connection between chordal distance for subspacesand the distance between the projections onto these subspaces. This resultcan be found in [29] in the language of Hilbert-Schmidt norms.

28 Peter G. CasazzaProposition 7. Let H M be an M-dimensional Hilbert space with orthonormalbasis {e i } M i=1 . Let P, Q be the orthogonal projections of HM onto N-dimensional subspaces W 1 , W 2 respectively. Then the chordal distance betweenW 1 , W 2 satisfiesd 2 c(W 1 , W 2 ) = 1 M∑‖P e i − Qe i ‖ 2 .2i=1In particular, there are orthonormal bases {e i } N i=1 for W 1 and {ẽ i } N i=1 for W 2satisfying1M∑N∑N∑‖P e i − Qe i ‖ 2 ≤ ‖e i − ẽ i ‖ 2 ≤ 2 ‖P e i − Qe i ‖ 2 .2i=1Proof. We compute:i=1i=1i=1M∑M∑‖P e i − Qe i ‖ 2 = 〈P e i − Qe i , P e i − Qe i 〉i=1M∑M∑M∑= ‖P e i ‖ 2 + ‖Qe i ‖ 2 − 2 〈P e i , Qe i 〉i=1= 2N − 2i=1M∑〈P Qe i , e i 〉i=1= 2N − 2T r P Q= 2N − 2[N − d 2 c(W 1 , W 2 )]= 2d 2 c(W 1 , W 2 ).This combined with Equation 1.5 completes the proof.The next problem to be addressed is to connect the distance betweenprojections and the distance between the corresponding ranges of analysisoperators for Parseval frames.Theorem 16. Let P, Q be projections of rank N on H M and let {e i } M i=1 bethe coordinate basis of H M . Further, assume that there is a Parseval frame{ϕ i } M i=1 for HN with analysis operator T satisfying T ϕ i = P e i for all i =1, 2, . . . , M. IfM∑‖P e i − Qe i ‖ 2 < ɛ,i=1then there is a Parseval frame {ψ i } M i=1 for H M with analysis operator T 1satisfyingT 1 ψ i = Qe i , for all i = 1, 2, . . . , M,andi=1

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 29M∑‖ϕ i − ψ i ‖ 2 < 2ɛ.i=1Moreover, if {Qe i } M i=1 is equal norm, then {ψ i} M i=1 may be chosen to be equalnorm.Proof. By Proposition 7, there are orthonormal bases {a j } M j=1 and {b j} M j=1for W 1 = T (H N ), W 2 = T 1 (H N ) respectively satisfyingM∑‖a j − b j ‖ 2 < 2ɛ.j=1Let A, B be the N ×M matrices whose j th columns are a j , b j respectively. Leta ij , b ij be the (i, j) entry of A, B respectively. Finally, let {ϕ ′ i }M i=1 , {ψ′ i }M i=1 bethe i th rows of A, B respectively. Then we haveM∑M∑ N∑‖ϕ ′ i − ψ i‖ ′ 2 = |a ij − b ij | 2i=1==i=1 j=1N∑j=1 i=1M∑|a ij − b ij | 2N∑‖a j − b j ‖ 2i=1≤ 2ɛ.Since the columns of A form an orthonormal basis for W 1 , we know that{ϕ ′ i }M i=1 is a Parseval frame which is isomorphic to {ϕ i} M i=1 . Thus there isa unitary operator U : H M → H M with Uϕ ′ i = ϕ i. Now let {ψ i } M i=1 ={Uψ i ′}M i=1 . ThenM∑M∑M∑‖ϕ i − Uψ i‖ ′ 2 = ‖U(ϕ ′ i) − U(ψ i)‖ ′ 2 = ‖ϕ ′ i − ψ i‖ ′ 2 ≤ 2ɛ.i=1i=1Finally, if T 1 is the analysis operator for the Parseval frame {ψ i } M i=1 , thenT 1 is a isometry and since T 1 ψ i = Qe i , for all i = 1, 2, . . . , N, if Qe i is equalnorm, so is {T 1 ψ i } M i=1 and hence so is {ψ i} N i=1 .Theorem 17. If g(ɛ, N, M) is the function for the Paulsen Problem andf(ɛ, N, M) is the function for the Projection Problem, theni=1f(ɛ, N, M) ≤ 4g(ɛ, N, M) ≤ 8f(ɛ, N, M).Proof. First, assume that the Projection Problem holds with function f(ɛ, N, M).Let {ϕ i } M i=1 be a Parseval frame for HN satisfying

30 Peter G. Casazza(1 − ɛ) N M ≤ ‖ϕ i‖ 2 ≤ (1 + ɛ) N M .Let T be the analysis operator of {ϕ i } M i=1 and let P be the projection of H Monto range T . So, T ϕ i = P e i for all i = 1, 2, . . . , M. By our assumption thatthe Projection Problem holds, there is a projection Q on H M with constantdiagonal so thatM∑‖P e i − Qe i ‖ 2 ≤ f(ɛ, N, M).i=1By Theorem 16, there is a a Parseval frame {ψ i } M i=1 for HN with analysisoperator T 1 so that T 1 ψ i = Qe i andM∑‖ϕ i − ψ i ‖ 2 ≤ 2f(ɛ, N, M).i=1Since T 1 is a isometry and {T 1 ψ i } M i=1 is equal norm, it follows that {ψ i} M i=1is an equal norm Parseval frame satisfying the Paulsen problem.Conversely, assume the Parseval Paulsen problem has a positive solutionwith function g(ɛ, N, M). Let P be an orthogonal projection on H M satisfying(1 − ɛ) N M ≤ ‖P e i‖ 2 ≤ (1 + ɛ) N M .Then {P e i } M i=1 is a ɛ-nearly equal norm Parseval frame for HN and by thePaulsen problem, there is an equal norm Parseval frame {ψ i } M i=1 so thatM∑‖ϕ i − ψ i ‖ 2 < g(ɛ, N, M).i=1Let T 1 be the analysis operator of {ψ i } M i=1 . Letting Q be the projection ontothe range of T 1 , we have that Qe i = T 1 ψ i , for all i = 1, 2, . . . , M. By Theorem15, we have thatM∑N∑‖P e i − T 1 ψ i ‖ 2 = ‖P e i − Qe i ‖ 2 ≤ 4g(ɛ, N, M).i=1i=1Since T 1 is a isometry and {ψ i } M i=1constant diagonal projection.is equal norm, it follows that Q is aIn [14] there are several generalizations of the Paulsen and ProjectionProblems.

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 311.5 Final CommentsWe have concentrated here on some problems in pure mathematics which havefinite dimensional formulations. There are many other infinite dimensionalversions of these problems [26, 27] in sampling theory, harmonic analysis andmore which we have not covered.Because of the long history of these problems and their connections to somany areas of mathematics, we are naturally led to consider the decidabilityof KS. Since we have finite dimensional versions of the problem, it can bereformulated in the language of pure number theory, and hence it has aproperty logicians call absolutness. As a practical matter, the general feelingis that this means it is very unlikely to be undecidable.AcknowledgementsThe author acknowledges support from NSF DMS 1008183, NSF ATD1042701and AFOSR DGE51: FA9550-11-1-0245.References1. C.A. Akemann and J. Anderson, Lyapunov theorems for operator algebras, Memoirsof AMS 94 (1991).2. J. Anderson, Restrictions and representations of states on C ∗ -algebras, Transactionsof AMS 249 (1979) 303–329.3. J. Anderson, Extreme points in sets of positive linear maps on B(H), Journalof Functional Analysis 31 (1979) 195–217.4. J. Anderson, A conjecture concerning pure states on B(H) and a related theorem,in Topics in modern operator theory, Birkhäuser (1981) 27–43.5. R. Balan, Equivalence relations and distances between Hilbert frames. Proceedingsof AMS 127 (8) (1999), 2353-2366.6. R. Balan, P.G. Casazza, C. Heil and Z. Landau, Density, overcompleteness andlocalization of frames. I. Theory, Journal of Fourier Analysis and Applications,12 (2006) pp. 105-143.7. R. Balan, P.G. Casazza, C. Heil and Z. Landau, Density, overcompleteness andlocalization of frames. II. Gabor systems, Journal of Fourier Analysis and Applications,12 (2006) pp. 309-344.8. K. Berman, H. Halpern, V. Kaftal and G. Weiss, Matrix norm inequalities andthe relative Dixmier property, Integral Equations and Operator Theory 11 (1988)28–48.9. K. Berman, H. Halpern, V. Kaftal and G. Weiss, Some C 4 and C 6 norm inequalitiesrelated to the paving problem, Proceedings of Symposia in Pure Math.51 (1970) 29-41.10. B. Bodmann and P.G. Casazza, The road to equal-norm Parseval frames, Journalof Functional Analysis 258, No. 2 (2010) 397-420.

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