The Kadison-Singer and Paulsen Problems in Finite Frame Theory

Transcription

1 Chapter 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory Peter G. Casazza Abstract We now know that some of the basic open problems in frame theory are equivalent to fundamental open problems in a dozen areas of research in both pure and applied mathematics, engineering and more. This includes the 1959 Kadison-Singer Problem in C -algebras, the Paving Conjecture in operator theory, the Bourgain-Tzafriri Conjecture in Banach space theory, the Feichtinger Conjecture and the R ɛ -Conjecture in frame theory and much more. In this chapter we will show these equivalences and many more. We will also consider a slight weakening of the Kadison-Singer Problem called the Sundberg Problem. Then we will look at the recent advances on another deep problem in frame theory called the Paulsen Problem. In particular, we will see that this problem is also equivalent to a fundamental open problem in operator theory. Namely, if a projection on a finite dimensional Hilbert space has a nearly constant diagonal, how close is it to a constant diagonal projection? Key words: Kadison-Singer Problem, Paving Conjecture, State, Rieszable, Discrete Fourier Transform, R ɛ -Conjecture, Feichtinger Conjecture, Bourgain-Tzafriri Conjecture, Restricted Invertibility Principle, Sundberg Problem, Paulsen Problem, Principle Angles, Chordal Distance 1.1 Introduction Finite frame theory is beginning to have an important impact on some of the deepest problems in both pure and applied mathematics. In this chapter we will look at two cases where finite frame theory is having a serious impact: Peter G. Casazza University of Missouri, Mathematics Department, Columbia, MO 6511, USA, casazzap@missouri.edu 1

2 Peter G. Casazza The Kadison-Singer Problem and the Paulsen Problem. We warn the reader that because we are restricting ourselves to finite dimensional Hilbert spaces, a significant body of literature on the infinite dimensional versions of these problems does not appear here. 1. The Kadison-Singer Problem For over 50 years the Kadison-Singer Problem [36] has defied the best efforts of 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, the Hilbert space of square summable sequences on the integers, have a unique extension to a (pure) state on B(l ), i.e., the von Neumann algebra of all bounded linear operators on the Hilbert space l? A state of a von Neumann algebra R is a linear functional f on R for which f(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 is compact in the ω -topology. By the Krein-Milman theorem, this convex set is the closed convex hull of its extreme points. The extremal elements in the space of states are called the pure states (of R). This problem arose from the very productive collaboration of Kadison and Singer in the 1950s when they were studying Dirac s Quantum Mechanics book [30] which culminated in their seminal work on triangular operator algebras. It is now known that the 1959 Kadison-Singer Problem is equivalent to fundamental unsolved problems in a dozen areas of research in pure mathematics, applied mathematics and engineering (See [1,, 3, 4, 17, 6, 7, 37] and their references). We will not develop this topic in detail here since it is fundamentally an infinite dimensional problem and we are concentrating on finite dimensional frame theory. In this chapter we will look at a number of these finite dimensional problems which are equivalent to KS and which are impacted by finite frame theory. Most people today seem to agree with the original statement of Kadison and Singer [36] that KS will have a negative answer and so all the equivalent forms will have negative answers also The Paving Conjecture A significant advance on KS was made by Anderson [] in 1979 when he reformulated KS into what is now known as the Paving Conjecture (See also [3, 4]). Lemma 5 of [36] shows a connection between KS and Paving. For

3 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 3 notation, if J {1,,..., n}, the diagonal projection Q J is the matrix whose entries are all zero except for the (i, i) entries for i J which are all one. 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 ) is said to have an (r, ɛ)-paving if there is a partition {A j } r j=1 of {1,,..., N} so that Q Aj T Q Aj ɛ T. Paving Conjecture 1 (PC) For every 0 < ɛ < 1, there is a natural number r so that for every natural number N and every linear operator T on l N whose 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 arbitrary operator T satisfies PC if T D(T ) satisfies PC where D(T ) is the diagonal of T. The only large classes of operators which have been shown to be pavable are diagonally dominant matrices [6, 7, 11, 31], l 1 -localized operators [4], matrices with all entries real and positive [3], matrices with small coefficients in comparison with the dimension [13] (See [40] for a paving into blocks of constant size), and Toeplitz operators over Riemann integrable functions (See also [33]). Also, in [9] there is an analysis of the paving problem for certain Schatten C p -norms. Theorem 1. The Paving Conjecture has a positive solution if any one of the following classes satisfies the Paving Conjecture: 1. Unitary operators. [7]. Orthogonal projections. [7] 3. Orthogonal projections with constant diagonal 1/. [18] 4. Positive operators. [7] 5. Self-adjoint operators. [7] 6. Gram matrices ( ϕ i, ϕ j ) i,j I where T : l (I) l (I) is a bounded linear operator, and T e i = ϕ i, T e i = 1 for all i I. [7] 7. Invertible operators (or invertible operators with zero diagonal). [7] 8. Triangular operators [38] Recently, Weaver [43] provided important insight into KS by giving an equivalent problem to PC in terms of projections. Conjecture 1 (Weaver). There exist universal constants 0 < δ, ɛ < 1 and r N so that for all N and all orthogonal projections P on l N with δ(p ) δ, there is a paving {A j } r j=1 of {1,,..., N} so that Q A j P Q Aj 1 ɛ, for all j = 1,,..., r. This needs some explanation since there is nothing in [43] that looks anything like Conjecture 1. Weaver observes that the fact that Conjecture 1

4 4 Peter G. Casazza implies 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 Weaver shows 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 =, even for projections with constant diagonal 1/. Recently [1] there appeared a frame theoretic concrete construction of non--pavable projections. If this construction can be generalized, we would have a counterexample to PC and KS. We now look at the construction from [1]. Definition. A family of vectors {ϕ i } M for an N-dimensional Hilbert space H N is (δ, r)-rieszable if there is a partition {A j } r j=1 of {1,,..., M} so that for all j = 1,,..., r and all scalars {a i } i Aj we have a i ϕ i δ a i. A projection P on H N is (δ, r)-rieszable if {P e i } N is (δ, r)-rieszable. We now have: Proposition 1. Let P be an orthogonal projection on H N. The following are equivalent: (1) The vectors {P e i } N are (δ, r)-rieszable. () There is a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all scalars {a i } i Aj we have a i (I P )e i (1 δ) a i. (3) The matrix of I P is (δ, r)-pavable. Proof. (1) (): For any scalars {a i } i AJ we have a i = a i P e i + a i (I P )e i. Hence, a i (I P )e i (1 δ) a i if and only if a i P e i δ a i. () (3): Given any partition {A j } r j=1, any 1 j r and any x = a i e i we have

5 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 5 (I P )x, x = (I P )x = a i (I P )e i if and only if I P (1 δ)i. (1 δ) a i = (1 δ)x, x, Given N N, let ω = exp ( πi N ), we define the discrete Fourier transform (DFT) matrix in C N by 1 ( D N = ω jk ) N 1 N j,k=0. The main point of these D N matrices is that they are unitary matrices 1 for which the modulus of all the entries are equal to N. The following is a simple observation. Proposition. Let A = (a ij ) N i,j=1 be a matrix with orthogonal rows and satisfying a ij = a for all i, j. If we multiply the j th -row of A by a constant C j to get a new matrix B, then: (1) The rows of B are orthogonal. () The square sums of the entries of any column of B all equal a N Cj. j=1 (3) The square sum of the entries of the j th row of B equals ac j. To construct our example, we start with a N N DFT and multiply the first N 1 rows by and the remaining rows by N+1 to get a new matrix B 1. Next, we take a second N N DFT matrix and multiply the N first N 1 rows by 0 and the remaining rows by N+1 to get a matrix B. We then put the matrices B 1, B side by side to get a N N matrix B of the form B = (N-1) Rows 0 (N+1) Rows N+1 N N+1 This matrix has N rows and 4N columns. Now we show that this matrix gives the required example. Proposition 3. The matrix B satisfies: (1) The columns are orthogonal and the square sum of the coefficients of every column equals.

6 6 Peter G. Casazza () The square sum of the coefficients of every row equals 1. The row vectors of the matrix B are not (δ, )-Rieszable, for any δ independent of N. Proof. A direct calculation yields (1) and (). We will now show that the column vectors of B are not uniformly two Rieszable independent of N. So let {A 1, A } be a partition of {1,,..., 4N}. Without loss of generality, we may assume that A 1 {1,,..., N} N. Let the column vectors of the matrix B be {ϕ i } 4N as elements of CN. Let P N 1 be the orthogonal projection of C N onto the first N 1 coordinates. Since A 1 N, there are scalars {a i } i A1 so that i A 1 a i = 1 and P N 1 ( i A 1 a i ϕ i ) = 0. Also, let {ψ j } N j=1 be the orthonormal basis consisting of the original columns of the DF T N. We now have: a i ϕ i = (I P N 1 )( a i ϕ i ) i A 1 i A 1 = N + 1 (I P N 1)( N + 1 = N + 1 = N + 1. a i ψ i i A 1 a i i A 1 i A 1 a i ψ i ) Letting N, this class of matrices is not (δ, )-Rieszable, and hence not (δ, )-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 to PC. 1.. The R ɛ -Conjecture In this section we will define the R ɛ -Conjecture and show it is equivalent to the Paving Conjecture. Definition 3. A family of vectors {ϕ i } M is an ɛ-riesz basic sequence for 0 < ɛ < 1 if for all scalars {a i } M we have

7 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 7 (1 ɛ) a i a i ϕ i (1 + ɛ) a i. A natural question is whether we can improve the Riesz basis bounds for a unit norm Riesz basic sequence by partitioning the sequence into subsets. Conjecture (R ɛ -Conjecture). For every ɛ > 0, every unit norm Riesz basic sequence is a finite union of ɛ-riesz basic sequences. This conjecture was first stated by Casazza and Vershynin and was first studied in [16] where it was shown that PC implies the conjecture. One advantage of the R ɛ -Conjecture is that it can be shown to students at the beginning 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 ) with T e i = 1 for i = 1,,..., N there is an r = r(ɛ, T ) and a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all scalars {a i } i Aj we have (1 ɛ) a i a i T e i (1 + ɛ) a i. Now we show that the R ɛ -Conjecture is equivalent to PC. Theorem. The following are equivalent: (1) The Paving Conjecture. () For 0 < ɛ < 1, there is an r = r(ɛ, B) so that for every N N, if T : l N l N is a bounded linear operator with T B and T e i = 1 for all i = 1,,..., N, then there is a partition {A j } r j=1 of {1,,..., N} so that for each 1 j r, {T e i } i Aj is an ɛ-riesz basic sequence. (3) The R ɛ -Conjecture. Proof. (1) (): Fix 0 < ɛ < 1. Given T as in (), let S = T T. Since S has ones on its diagonal, by the Paving Conjecture there is a r = r(ɛ, T ) and a partition {A j } r j=1 of {1,,..., N} so that for every j = 1,,..., r we have where δ = Q Aj (I S)Q Aj δ I S ɛ S +1. Now, for all x = N a ie i and all j = 1,,..., r we have a i T e i = T Q Aj x = 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 δ I S Q Aj x (1 ɛ) Q Aj x = (1 ɛ) a i.

8 8 Peter G. Casazza Similarly, a i T e i (1 + ɛ) a i. () (3): This is obvious. (3) (1): Let T B(l N ) 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 is pavable. Fix 0 < δ < 1 and let ɛ > 0. Let ψ i = 1 δ ϕ i δe i l N l N. Then ψ i = 1 for all 1 i N and for all scalars {a i } N N N N δ a i a i ψ i = (1 δ ) a i T e i + δ [ (1 δ ) T + δ ] N a i. N a i So {ψ i } N is a unit norm Riesz basic sequence and ψ i, ψ k = (1 δ ) 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,,..., N} so that for all j = 1,,..., r and all x = a i e i, (1 ɛ) a i a i ψ i = a i ψ i, a k ψ k k A j = a i ψ i + a i a k ψ i, ψ k i k A j = a i + (1 δ ) a i a k ϕ i, ϕ k i k A j = a i + (1 δ ) Q Aj (G D(G))Q Aj x, x (1 + ɛ) a i. Subtracting a i through the inequality yields, That is, ɛ a i (1 δ ) Q Aj (G D(G))Q Aj x, x ɛ a i. (1 δ ) Q Aj (G D(G))Q Aj x, x ɛ x. Since Q Aj (G D(G))Q Aj is a self-adjoint operator, we have (1 δ ) Q Aj (G D(G))Q Aj ɛ, i.e., (1 δ )G (and hence G) is pavable. Remark 1. The proof of (3) (1) of Theorem illustrates a standard method for turning conjectures about unit norm Riesz basic sequences {ψ i } i I into conjectures about unit norm familes {ϕ i } i I with T B(l (I)) and T e i = ϕ i. Namely, given {ϕ i } i I and 0 < δ < 1 let ψ i = 1 δ f i δe i l (I) l (I).

9 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 9 Then {ψ i } i I is a unit norm Riesz basic sequence and for δ small enough, ψ i is 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 chapter in that it does not hold for equivalent norms on the Hilbert space in general. For example, if we renorm l by: {a i } = {a i } l + sup i a i then the R ɛ - Conjecture fails for this equivalent norm. To see this, we proceed by way of contradiction and assume there is an 0 < ɛ < 1 and an r = r(ɛ, ) satisfying the R ɛ -Conjecture. Let {e i } N be the unit vectors for ln and let x i = e i+e +1 i+1 for 1 i N. This is now a unit norm Riesz basic sequence with upper Riesz bound. Assume we partition {1,,..., N} 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 r and a i = 1 N for i A. Then ( ) 1 r a i x i = + i A + 1 N Since the norm above is bounded away from one for large N, we cannot satisfy the requirements of the R ɛ -Conjecture. It follows that a positive solution to KS would imply a fundamental new result concerning inner products, not just norms. Another important equivalent form of PC comes from [6]. This is, on its face, a significant weakening of the R ɛ -Conjecture while it still remains equivalent to PC. Conjecture 4. There exists a constant A > 0 and a natural number r so that for all natural numbers N and all T : l N l N with T e i = 1 for all i = 1,,..., N and T, there is a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all scalars {a i } i Aj we have a i T e i A a i. Theorem 3. Conjecture 4 is equivalent to PC. Proof. Since PC is equivalent to the R ɛ -Conjecture, which in turn implies Conjecture 4, we just need to show that Conjecture 4 implies Conjecture 1. So choose r, A satisfying Conjecture 4. Fix 0 < δ 3 4 and let P be an orthogonal projection on l N with δ(p ) δ Now, P e i, e i = P e i δ implies (I P )e i 1 δ 1 4. Define T : ln l N by T e i = For any scalars {a i } N we have (I P )ei (I P )e i.

10 10 Peter G. Casazza N a i T e i N a i = (I P )e i (I P )e i N a i (I P )e i 4 N a i. So T e i = 1 and T. By Conjecture 4, there is a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all scalars {a i } i Aj we have a i T e i A a i. i A J Hence, a i (I P )e i = a i (I P )e i T e i A a i (I P )e i A a i. 4 It follows that for all scalars {a i } i Aj, a i = a i P e i + a i (I P )e i a i P e i + A a i. 4 Now, for all x = N a ie i we have P Q Aj x = a i P e i (1 A 4 ) a i. Thus, Q Aj P Q Aj = P Q Aj 1 A 4. So Conjecture 1 holds. Weaver [43] established an important relationship between frames and PC by showing that the following conjecture is equivalent to PC.

11 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 11 Conjecture 5. There are universal constants B 4 and α > B and an r N so that the following holds. Whenever {ϕ i } M is a unit norm B-tight frame for l N, there exists a partition {A j } r j=1 of {1,,..., M} so that for all j = 1,,..., r and all x l N we have x, ϕ i (B α) x. (1.1) Using Conjecture 5 we can show that the following conjecture is equivalent to PC: Conjecture 6. There is a universal constant 1 D so that for all T B(l N ) with T e i = 1 for all i = 1,,..., N, there is an r = r( T ) and a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all scalars {a i} i Aj a i T e i D a i. Theorem 4. Conjecture 6 is equivalent to PC. Proof: Since Conjecture 3 clearly implies Conjecture 6, we just need to show that Conjecture 6 implies Conjecture 5. So, choose D as in Conjecture 6 and choose B 4 and α > B so that D B α. Let {ϕ i } M be a unit norm B tight frame for l N. If T e i = ϕ i is the synthesis operator for this frame, then T = T = B. So by Conjecture 6, there is an r = r( B ) and a partition {A j } r j=1 of {1,,..., M} so that for all j = 1,,..., r and all scalars {a i } i Aj a i T e i = a i ϕ i D a i (B α) a i. So T Q Aj B α and for all x l N x, ϕ i = (Q Aj T ) x T Q Aj x (B α) x. This verifies that Conjecture 5 holds and so KS holds. Remark 1 and Conjecture 6 show that we only need any universal upper bound in the R ɛ -Conjecture to hold for KS The Feichtinger Conjecture While working in time-frequency analysis, Feichtinger [16] observed that all of the Gabor frames he was using had the property that they could be divided

12 1 Peter G. Casazza into a finite number of subsets which were Riesz basic sequences. This led to the 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) > 0 so that whenever {ϕ i } N is a frame for HN with upper frame bound B and ϕ i C for all i = 1,,..., N, then {1,,..., N} can be partitioned into subsets {A j } r j=1 so that for each 1 j r, {ϕ i} i Aj is a Riesz basic sequence with 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, it remains 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. () The Feichtinger Conjecture. Proof. (1) (): Part () of Theorem is equivalent to PC and clearly implies FC. () (1): We will observe that FC implies Conjecture 4 which is equivalent to PC by Theorem 3. In Conjecture 4, {T e i } N is a frame for its span with upper frame bound. It is now immediate that the Finite Dimensional Feichtinger 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 following holds. Let {ϕ i } M be elements of ln with ϕ i 1 for i = 1,,..., M and suppose for every x l N, x, ϕ i B x. (1.) Then, there is a partition {A j } r j=1 of {1,,..., M} so that for all x ln all j = 1,,..., r, x, ϕ i (B ɛ) x. and Theorem 6. The following are equivalent: (1) The Paving Conjecture. () Conjecture KS r holds for some r.

13 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 13 Proof. Assume Conjecture KS r is true for some fixed r, B, ɛ. We will show that Conjecture 1 is true. Let P be an orthogonal projection on H M with δ(p ) 1 B. If P has rank N, then its range is a N-dimensional subspace W of H M. Define ϕ i = B P e i W for all 1 i M. We check ϕ i = B P e i = B P e i, e i Bδ(P ) 1, for all i = 1,,..., M. Now, if x W is any unit vector then x, ϕ i = x, BP e i = B x, e i = B. By Conjecture KS r, there is a partition {A j } r j=1 of {1,,..., M} satisfying for all 1 j r and all unit vectors x W, x, ϕ i B ɛ. Then r j=1 Q A j = Id, and for any unit vector x W we have Q Aj P x = Q Aj P x, e i = x, P Q j e i = 1 x, ϕ i ɛ B B. Thus Conjecture 1 holds and so PC holds. Conversely, assume KS r fails for all r. Fix B = r and let {ϕ i } M in H N be a counterexample with ɛ = 1. Let ψ i = ϕi B and note that ψ i ψ T i = ψ i 1 B, for all i = 1,,..., M and M ψ iψ T i Id. Then Id M ψ iψi T is a positive finite rank operator, so we can find positive rank one operators ψ i ψi T for M + 1 i K such that ψ i ψi T 1 B for all 1 i K and K ψ iψi T = Id. Let T be the analysis operator for {ψ i } K, which is an isometry and if P is the orthogonal projection of H K with range T (H N ), then P e i = T ψ i for all i = 1,,..., K. Let D be the diagonal matrix with the same diagonal as P. Then D = max ψ i 1 1 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,,..., K by letting A j be the diagonal of Q j. By our choice of {ϕ i } M, there exists 1 j r and x HN with x = 1 and

14 14 Peter G. Casazza x, ϕ i > B 1. Hence, It follows that for all j, Q j P (T x) {1,,...,M} x, ψ i > 1 1 B. O Q j P (T x), e i = = x, ψ i > 1 1 B. T x, e i Thus, Q j P Q j = Q j P > 1 1 B. Now, the matrix A = P D has zero diagonal and satisfies A B and the above shows that for any K K diagonal projections {Q j } r j=1 with r j=1 Q j = Id we have Q j AQ j Q j P Q j Q j DQ j 1, for some j. B Finally, as B = r, we obtain a sequence of examples which negate the Paving Conjecture. Weaver [43] also shows that Conjecture KS r is equivalent to PC if we assume equality in Equation 1. for all x l M. Weaver further shows that Conjecture KS r is equivalent to PC even if we strengthen its assumptions so as to require that the vectors {ϕ i } M are of equal norm and that equality holds in 1., but at great cost to our ɛ > 0. Conjecture 9 (KS r ). There exists universal constants B 4 and ɛ > B so that the following holds. Let {ϕ i } M be elements of ln with ϕ i = 1 for i = 1,,..., M and suppose for every x l N, x, ϕ i = B x. (1.3) Then, there is a partition {A j } r j=1 of {1,,..., M} so that for all x lm all j = 1,,..., r, x, ϕ i (B ɛ) x. and We introduce one more conjecture. Conjecture 10. There exist universal constants 0 < δ, δ ɛ < 1 and r N so that for all N and all orthogonal projections P on l N with δ(p ) δ

15 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 15 and P e i = P e j for all i, j = 1,,..., N, there is a paving {A j } r j=1 of {1,,..., N} so that Q Aj P Q Aj 1 ɛ, for all j = 1,,..., 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)) implies Conjecture 10. So we assume that Conjecture 10 holds and we will show that Conjecture 9 holds. Let {ϕ i } M be elements of HN with ϕ i = 1 for i = 1,,..., M and suppose for every x H N, x, ϕ i = B x, (1.4) where 1 1 B δ. It follows from Equation 1.4 that { B ϕ i } M is an equal norm Parseval frame and so by Naimark s Theorem, we may assume there is a larger Hilbert space l M and a projection P : l M H N so that P e i = ϕ i for all i = 1,,..., M. Now P e i = P e i, e i = 1 B δ for all i = 1,,..., N. So by Conjecture 10, there is a paving {A j } r j=1 of {1,,..., M} so that Q Aj P Q Aj 1 ɛ, for all j = 1,,..., r. Now for all 1 j r and all x l N we have: It follows that for all x l N Q Aj P x = = Q Aj P x, e i x, P Q Aj e i = 1 x, ϕ i B we have Q Aj P x = Q Aj P Q Aj x (1 ɛ) x. xϕ i (B ɛb) x. Since ɛb > B, we have verified Conjecture 9.

16 16 Peter G. Casazza 1..4 The Bourgain-Tzafriri Conjecture We start with a fundamental theorem of Bourgain and Tzafriri called the Restricted Invertibility Principle. This theorem led to the (strong and weak) Bourgain-Tzafriri Conjectures. We will see that these conjectures are equivalent to PC. In 1987, Bourgain and Tzafriri [1] proved a fundamental result in Banach space theory known as the Restricted Invertibility Principle. Theorem 8 (Bourgain-Tzafriri). There is a universal constants 0 < c < 1 so that whenever T : l N l N is a linear operator for which T e i = 1, for 1 i N, then there exists a subset σ {1,,..., N} of cardinality σ cn/ T so that for all choices of scalars {a j } j σ, a j T e j c j σ j σ a j. A close examination of the proof of the theorem [1] yields that c is on the order of The proof of the theorem uses probabilistic and function analytic techniques, is non-trivial and is non-constructive. A significant breakthrough occurred recently when Spielman and Srivastava [39] presented an algorithm for proving the Restricted-Invertibility Theorem. Moreover, their proof gives the best possible constants in the theorem. Theorem 9 (Restricted Invertibility Theorem: Spielman-Srivastava Form). Assume {v i } M are vectors in ln with A = M v ivi T = I and 0 < ɛ < 1. If L : l N l N is a linear operator, then there is a subset J {1,,..., M} of size J ɛ L F L and ( ) λ min Lv i (Lv i ) T i J for which {Lv i } i J is linearly independent > (1 ɛ) L F M where L F is the Frobenious norm of L and λ min is the smallest eigenvalue of the operator computed on span {v i } i J. This generalized form of the Restricted-Invertibility Theorem was introduced by Vershynin [41] where he studied the contact points of convex bodies using John s decompositions of the identity. The corresponding theorem for infinite dimensional Hilbert spaces is still open. But this case requires the set J to be large with respect to the Beurling density [6, 7]. Special cases of this problem were solved in [4, 41]. The inequality in the Restricted Invertibility Theorem is referred to as a lower l -bound. It is known [8, 41] that there is a corresponding close to one upper l -bound which can be achieved in the theorem. Recently, a two-sided Spielman-Srivastava algorithm was given [5] which gives the best two sided lower and upper bounds in BT.,

17 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 17 Corollary 1. For T : l N l N and an orthonormal basis {e i } N with T e i = 1, i = 1,..., N, and for any α (0, 1) there exists a subset J α {1,,..., N}, satisfying α N 1. J α T, and. 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 still open. Special cases of this problem were solved in [4, 41]. Theorem 8 gave rise to a problem in the area which has received a great deal of attention [13, 6, 7]. 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: For any natural number N, if T : l N l N is a linear operator with T B and T e i = 1 for all i = 1,,..., N, then there is a partition {A j } r j=1 of {1,,..., N} so that for all j = 1,,..., r and all choices of scalars {a i } i Aj we have: a i T e i A a i. Sometimes BT is called strong BT since there is a weakening called weak BT. In weak BT we allow A to depend upon the norm of the operator T. A significant amount of effort over the years was invested in trying to show that strong and weak BT are equivalent. Casazza and Tremain finally proved this equivalence [6]. We will not have to do any work here since we developed all of the needed results in earlier sections. Theorem 10. The following are equivalent: (1) The Paving Conjecture. () The Bourgain-Tzafriri Conjecture. (3) The (weak) Bourgain-Tzafriri Conjecture. Proof. (1) () (3): The Paving Conjecture is equivalent to the R ɛ - Conjecture which clearly implies the Bourgain-Tzafriri Conjecture and this immediately implies the (weak) Bourgain-Tzafriri Conjecture. (3) (1): The weak Bourgain-Tzafriri Conjecture immediately implies Conjecture 4 which is equivalent to the Paving Conjecture Partitioning Frames into Frame Subsets A natural and frequently occurring problem in frame theory is to partition a frame into subsets each of which has good frame bounds. This innocent

18 18 Peter G. Casazza looking question turns out to be much deeper than it looks, and as we will now see, it is equivalent to PC. Conjecture 11. There exists an ɛ > 0 so that for large K, for all N and all equal norm Parseval frames {ϕ i } KN for ln, there is a J {1,,..., KN} so that both {ϕ i } i J and {ϕ i } i J c have lower frame bounds which are greater than ɛ. The ideal situation would be for Conjecture 11 to hold for all K. In order for {ϕ i } i J and {ϕ i } i J c to both be frames for l N, they at least have to span l N. So the first question is whether we can partition our frame into spanning 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+L, 0 L < N for l N can be partitioned into K linearly independent spanning sets plus a linearly independent set of L elements. The natural question is whether we can do such a partition so that each of the subsets has good frame bounds, i.e., a universal lower frame bound for all subsets. Before addressing this question, we state another conjecture. Conjecture 1. There exists ɛ > 0 and a natural number r so that for all N, all large K and all equal norm Parseval frames {ϕ i } KN in ln there is a partition {A j } r j=1 of {1,,..., KN} so that for all j = 1,,..., r the Bessel bound 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 1. () Conjecture 1 is equivalent to PC. Proof. (1): Fix ɛ > 0, r, K as in Conjecture 11. Let {ϕ i } KN be an equal norm Parseval frame for an N-dimensional Hilbert space H N. By Naimark s Theorem we may assume there is an orthogonal projection P on l KN with P e i = ϕ i for all i = 1,,..., KN. By Conjecture 11, there is a J {1,,..., KN} so that {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 KN ), KN x = x, P e i = x, P e i + x, P e i i J i J c i J x, P e i + ɛ x. That is, i J x, P e i (1 ɛ) x. 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 =

19 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 19 P Q J = (P Q J ) 1 ɛ. Similarly, Q J cp Q J c 1 ɛ. So Conjecture 1 holds for r =. (): We will show that Conjecture 1 implies Conjecture 5. Choose ɛ and r satisfying Conjecture 1 for all large K. In particular, choose any K with 1 K < α. Let {ϕ i } M be a unit norm K-tight frame for an N-dimensional Hilbert space H N. Then M = M ϕ i = KN. Since { 1 K ϕ i } M is an equal norm Parseval frame, by Naimark s Theorem we may assume there is an orthogonal projection P on l M with P e i = 1 K ϕ i, for i = 1,,..., M. By Conjecture 1 there is a partition {A j } r j=1 of {1,,..., M} so that the Bessel bound (P Q Aj ) for each family {ϕ i } i Aj is 1 ɛ. So for j = 1,,..., r and any x l N we have x, 1 ϕ i = K x, P Q Aj e i = Q Aj P x, e i Q Aj P x Q Aj P x = (P Q Aj ) x (1 ɛ) x. Hence, x, ϕ i K(1 ɛ) x = (K Kɛ) x. Since Kɛ > K, we have verified Conjecture 5. For the converse, choose r, δ, ɛ satisfying Conjecture 1. If {ϕ i } KN is an equal 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 projection P on l KN with P e i = ϕ i for i = 1,,..., KN. Since δ(p ) = ϕ i 1 K δ, by Conjecture 1 there is a partition {A j } r j=1 of {1,,..., KN} so that for all j = 1,,..., r, Q Aj P Q Aj = P Q Aj = (P Q Aj ) 1 ɛ. Since (P Q Aj ) is the Bessel bound for {P e i } i Aj that Conjecture 1 holds. = {ϕ i } i Aj, we have 1.3 The Sundberg Problem Recently, an apparent weakening of the Kadison-Singer Problem has arisen. In his work on interpolation in complex function theory, Sundberg noticed the following problem. Although this is an infinite dimensional problem, we state it here because of its connections to the Kadison-Singer Problem. Problem 1 (Sundberg Problem). If {ϕ} is a unit norm Bessel sequence, can we partition {ϕ i } into a finite number of spanning sets?

20 0 Peter G. Casazza This problem appears to be quite innocent, but it is surprisingly difficult. It is immediate that the Feichtinger Conjecture implies the Sundberg Problem. Theorem 1. A positive solution to the Feichtinger Conjecture implies a positive solution to the Sundberg Problem. Proof. If {ϕ i } is a unit norm Bessel sequence then by FC, we can partition the natural numbers into a finite number of sets {A j } r j=1 so that {ϕ i} i Aj is a Riesz sequence for all j = 1,,..., r. For each j = 1,,..., 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 Problem The Paulsen Problem has been intractable for over a dozen years despite receiving quite a bit of attention. In this section we look at the current state of the art on this problem. First we need two definitions. Definition 4. A frame {ϕ i } M for HN with frame operator S is said to be ɛ-nearly equal norm if (1 ɛ) N M ϕ i (1 + ɛ) N, for all i = 1,,..., M, M and it is ɛ-nearly Parseval if We also need: (1 ɛ)id S (1 + ɛ)id. Definition 5. Given frames Φ = {ϕ i } M and Ψ = {ψ i} M for HN, we define the distance between them by d(φ, Ψ) = ϕ i ψ i. The above is not exactly a distance function since we have not taken the square root of the right-hand-side of the equality. But since this formulation is standard, we will use it. We can now state the Paulsen Problem. Problem (Paulsen Problem). How close is an ɛ-nearly equal norm and ɛ-nearly Parseval frame to an equal norm Parseval frame? The importance of the Paulsen Problem is that we have algorithms for constructing frames which are equal norm and nearly Parseval. The question is, if we work with these frames, are we sure that we are working with a frame which is close to some equal norm Parseval frame? We are looking for the

21 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 1 function f(ɛ, N, M) so that every ɛ-nearly equal norm and ɛ-nearly Parseval frame Φ = {ϕ i } M satisfies d(φ, Ψ) f(ɛ, N, M), for some equal norm Parseval frame Ψ. A simple compactness argument due to 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 an 0 < ɛ so that for every δ = 1 n, there is a frame {ϕn i }M with frame bounds 1 1 n, n and satisfying ( 1 1 ) ( N n M ϕn i ) N n M, while Φ n = {ϕ n i }M is a distance greater than ɛ from any equal norm Parseval frame. By compactness and switching to a subsequence, we may assume that lim n ϕn i = ϕ i, exists for all i = 1,,..., M. But now Φ = {ϕ i } M is an equal norm Parseval frame contradicting the fact that d(φ n, Φ) ɛ > 0, for all n = 1,,..., for any equal norm Parseval frame. The problem with this argument is that it does not give any quantitative estimate on the parameters. We do not have a good idea of what form the function f(ɛ, N, M) must have. We do not even know if M needs to be in the function or if it is independent of the number of frame vectors. The following example shows that the Paulsen function is certainly a function of the dimension of the space. Lemma. The Paulsen function satisfies f(ɛ, N, M) ɛ N. Proof. Fix an ɛ > 0 and an orthonormal basis {e j } N j=1 for HN. We define a frame {ϕ i } N by ϕ i = { 1 ɛ e i 1+ɛ e i N if 1 i N if N + 1 i N By the definition, {ϕ i } M is ɛ-nearly equal norm. Also, for any x HN we have

22 Peter G. Casazza N x, ϕ i = (1 ɛ) N x, e i + (1 + ɛ) N x, e i = (1 + ɛ ) x. So {ϕ i } N is a (1 + ɛ ) tight frame and hence an ɛ-nearly Parseval frame. The closest equal norm frame to {ϕ i } N ei is { } N { ei } N. Also, N e i ϕ i + N i=n+1 e i N ϕ i = N ɛ e i + N ɛ e i = ɛ N. The main difficulty with solving the Paulsen Problem is that finding a close equal-norm frame to a given frame involves finding a close frame which satisfies a geometric condition while finding a close Parseval frame to a given frame involves satisfying a (algebraic) spectral condition. At this time, we lack techniques for combining these two conditions. However, each of them individually has a known solution. That is, we do know the closest equal norm frame to a given frame [15] and we do know the closest Parseval frame to a given frame [5, 10, 15,, 35]. Lemma 3. If {ϕ i } M is an ɛ-nearly equal norm frame in HN, then the closest equal norm frame to {ϕ i } M is where ψ i = a ϕ i, for i = 1,,..., M, ϕ i M a = ϕ i. M It is well known that for a frame {ϕ i } M for HN with frame operator S, the closest Parseval frame to {ϕ i } M is {S 1/ ϕ i } M [5, 10, 15,, 35]. We will give the version from [10] here. Proposition 5. Let {ϕ i } M be a frame for a N-dimensional Hilbert space H N, with frame operator S = T T. Then {S 1/ ϕ i } M is the closest Parseval frame to {ϕ i } M. Moreover, if {ϕ i} M is an ɛ-nearly Parseval frame then S 1/ ϕ i ϕ i N( ɛ 1 ɛ) Nɛ /4. Proof. We first check that {S 1/ ϕ i } M is the closest Parseval frame to {ϕ i } M. The squared l -distance between {ϕ i } M and any Parseval frame {ψ j} n j=1 can be expressed in terms of their analysis operators T and T 1 as F G = T r[(t T 1 )(T T 1 ) ] = T r[t T ] + T r[t 1 T 1 ] RT r[t T 1 ]

23 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 3 Choosing a Parseval frame {ψ i } M is equivalent to choosing the isometry T 1. To minimize the distance over all choices of T 1, consider the polar decomposition T = UP, where P is positive and U is an isometry. In fact, S = T T implies P = S 1/, 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 magnitude is bounded by the Cauchy Schwarz inequality, and thus it has a maximal real part if T 1 = U which implies T 1 U = I. In this case, T = T 1 P = T 1 S 1/, or equivalently T 1 = S 1/ T, and we conclude ψ i = S 1/ ϕ i for all i = 1,,... M. After choosing T 1 = T S 1/, the l -distance is expressed in terms of the eigenvalues {λ j } N j=1 of S = T T by F G = T r[s] + T r[i] T r[s 1/ ] N N = λ j + N λj. j=1 If 1 ɛ λ j 1 + ɛ for all j = 1,,... N, calculus shows that the maximum of λ j λ j is achieved when λ j = 1 ɛ. Consequently, j=1 F G N Nɛ N 1 ɛ. Estimating 1 ɛ by the first three terms in its decreasing power series gives the inequality F G Nɛ /4. It can be shown that the estimate above is exact and so we have separate verification that the closeness function is a function of N. There is a simple algorithm for turning any frame into an equal norm frame with the same frame operator due to Holmes and Paulsen [34]. Proposition 6. There is an algorithm for turning any frame into an equal norm frame with the same frame operator. Proof. Let {ϕ i } M be a frame for HN with frame operator S and analysis operator T. Then ϕ i = T r S. Let λ = T r S M. If ϕ i = λ, for all m = 1,,..., M, then we are done. Otherwise, there exists 1 i j M with ϕ i > λ > ϕ j. For any θ, replace the vectors ϕ i, ϕ j by the vectors ψ 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 is T 1 = UT for a unitary operator U on l N given by the Givens rotation. Hence, T1 T 1 = T U UT = T T = S; so

24 4 Peter G. Casazza the frame operator is unchanged for any value of θ. Now choose the θ yielding ψ i = λ. Repeating this process at most M 1 times yields an equal norm frame with the same frame operator as {ϕ i } M. Using a Parseval frame in Proposition 6, we do get to an equal norm Parseval frame. The problem, again, is that we do not have any quantitative measure of how close these two Parseval frames are. There is an obvious approach towards solving the Paulsen Problem. Given an ɛ-nearly equal norm ɛ-nearly Parseval frame {ϕ i } M for HN with frame operator S, we can switch to the closest Parseval frame {S 1/ ϕ i } M. Then switch to the closest equal norm frame to {S 1/ ϕ i } M, call it {ψ i} M with frame operator S 1. Now switch to {S 1/ 1 ψ i } M and again switch to the closest equal norm frame and continue. Unfortunately, even if we could show that this process converges and we could check the distance traveled through this process, we would still not have an answer to the Paulsen Problem because this process does not have to converge to an equal norm Parseval frame. In particular, there is a fixed point of this process which is not an equal norm Parseval frame. Example 1. Let {e i } N be an orthonormal basis for ln and let {ϕ i } N+1 be a equiangular unit norm tight frame for l N. Then {e i 0} N {0 ϕ i} N+1 in l N l N is a ɛ = 1 N -nearly equal norm and 1 N - nearly Parseval frame with frame operator, say S. A direct calculation shows that taking S 1/ of the frame vectors and switching to the closest equal norm frame leaves the frame unchanged. The Paulsen Problem has proven to be intractable for over 1 years. Recently, two partial solutions to the problem were given in [10, 0] each having some advantages. Since each of these papers is technical, we will only outline the ideas here. In [10], a new technique is introduced. This is a system of vector valued ODE s which starts with a given Parseval frame and has the property that all frames in the flow are still Parseval while approaching an equal norm Parseval frame. They then bound the arc length of the system of ODE s by the frame energy. Finally, giving an exponential bound on the frame energy, they have a quantitative estimate for the distance between the initial, ɛ-nearly equalnorm and 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 the Hilbert space and the number M of frame vectors are relatively prime. The authors show that in practice, this is not a serious restriction. The main result of [10] is Theorem 13. Let N, M N be relatively prime, let 0 < ɛ < 1, and assume Φ = {ϕ i } M is an ɛ-nearly equal-norm and ɛ-nearly Parseval frame for a real or complex Hilbert space of dimension N. Then there is an equal-norm Parseval frame Ψ = {ψ i } M such that

25 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 5 Φ Ψ 9 8 N M(M 1) 8 ɛ. In [0], the authors present a new iterative algorithm gradient descent of the frame potential for increasing the degree of tightness of any finite unit norm frame. The algorithm itself is trivial to implement, and it preserves certain group structures present in the initial frame. In the special case where the number of frame elements is relatively prime to the dimension of the underlying space, they show that this algorithm converges to a unit norm tight frame at a linear rate, provided the initial unit norm frame is already sufficiently close to being tight. The main difference between this approach and the approach in [10] is that in [10], the authors start with a nearly equal norm Parseval frame and improve its closeness to an equal norm frame while maintaining Parseval, and in [0] the authors start with an equal norm nearly Parseval frame and give an algorithm for improving its algebraic properties while changing its transform as little as possible. The main result from [0] is: 1 Theorem 14. Suppose M and N are relatively prime. Pick t (0, M ), and let Φ 0 = {ϕ i } M be a unit norm frame with analysis operator T 0 satisfying T0 T 0 M N I HS N. Now, iterate the gradient descent of the frame potential 3 method to obtain Φ k. Then, Φ := lim k Φ k exists and is a unit norm tight frame satisfying Φ Φ 0 HS 4N 0 M Mt T 0 T 0 M N I HS. In [10], the authors showed there is a connection between the Paulsen Problem and a fundamental open problem in operator theory. Problem 3 (Projection Problem). Let H N be an N-dimensional Hilbert space with orthonormal basis {e i } N. Find the function g(ɛ, N, M) satisfying the following. If P is a projection of rank M on H N satisfying (1 ɛ) M N P e i (1 + ɛ) M, for all i = 1,,..., N, N then there is a projection Q with Qe i = M N for all i = 1,,..., N satisfying N P e i Qe i g(ɛ, N, M). In [14], it was shown that the Paulsen problem is equivalent to the Projection Problem and that their closeness functions are within a factor of of one another. The proof of this result gives several exact connections between the distance between frames and the distance between the ranges of their analysis operators.

26 6 Peter G. Casazza Theorem 15. Let Φ = {ϕ i } i I, Ψ = {ψ i } i I be Parseval frames for a Hilbert space H with analysis operators T 1, T respectively. If d(φ, Ψ) = i I ϕ i ψ i < ɛ, then d(t 1 (Φ), T (Ψ)) = i I T 1 ϕ i T ψ i < 4ɛ. Proof. Note that for all j I, T 1 ϕ j = i I ϕ j, ϕ i e i, and T ψ j = i I ψ j, ψ i e i. Hence, T 1 ϕ j T ψ j = ϕ j, ϕ i ψ j, ψ i i I = ϕ j, ϕ i ψ i + ϕ j ψ j, ψ i i I ϕ j, ϕ i ψ i + ϕ j ψ j, ψ i. i I i I Summing over j and using the fact that our frames Φ and Ψ are Parseval gives T 1 ϕ j T ψ j ϕ j, ϕ i ψ i + ϕ j ψ j, ψ i j I j I i I j I i I = ϕ j, ϕ i ψ i + ϕ j ψ j i I j I j I = ϕ i ψ i + ϕ j ψ j i I j I = 4 ϕ j ψ j. j I Next, we want to relate the chordal distance between two subspaces to the distance between their orthogonal projections. First we need to define the distance between projections. Definition 6. If P, Q are projections on H N, we define the distance between them by d(p, Q) = P e i Qe i, where {e i } N is an orthonormal basis for HN.

27 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 7 The chordal distance between subspaces of a Hilbert space was defined by [9] and has been shown to have many uses over the years. Definition 7. Given M-dimensional subspaces W 1, W of a Hilbert space, define the M-tuple (σ 1, σ,..., σ M ) as follows: σ 1 = max{ x, y : x Sp W1, y Sp W } = x 1, y 1, where Sp W is the unit sphere of the subspace W. For i M, σ i = max{ x, y : x = y = 1, x j, x = 0 = y j, y, for 1 j i 1}, where σ i = x i, y i. The M-tuple (θ 1, θ,..., θ M ) with θ i = cos 1 (σ i ) is called the principle angles between W 1, W. The chordal distance between W 1, W is given by d c(w 1, W ) = sin θ i. So by the definition, there exists orthonormal bases {a j } M j=1, {b j} M j=1 W 1, W respectively satisfying ( ) θ a j b j = sin, for all j = 1,,..., M. for It follows that for 0 θ π, Hence, ( ) θ sin θ 4sin = a j b j 4sin θ, for all j = 1,,..., M. d c(w 1, W ) a j b j 4d c(w 1, W ). (1.5) j=1 We also need the following result [9]. Lemma 4. If H N is an N-dimensional Hilbert space and P, Q are rank M orthogonal projections onto subspaces W 1, W respectively, then the chordal distance d c (W 1, W ) between the subspaces satisfies d c(w 1, W ) = M T r P Q. Next we give a precise connection between chordal distance for subspaces and the distance between the projections onto these subspaces. This result can be found in [9] in the language of Hilbert-Schmidt norms.

28 8 Peter G. Casazza Proposition 7. Let H M be an M-dimensional Hilbert space with orthonormal basis {e i } M. Let P, Q be the orthogonal projections of HM onto N- dimensional subspaces W 1, W respectively. Then the chordal distance between W 1, W satisfies d c(w 1, W ) = 1 P e i Qe i. In particular, there are orthonormal bases {e i } N for W 1 and {ẽ i } N for W satisfying 1 N N P e i Qe i e i ẽ i P e i Qe i. Proof. We compute: P e i Qe i = P e i Qe i, P e i Qe i = P e i + Qe i P e i, Qe i = N P Qe i, e i = N T r P Q = N [N d c(w 1, W )] = d c(w 1, W ). This combined with Equation 1.5 completes the proof. The next problem to be addressed is to connect the distance between projections and the distance between the corresponding ranges of analysis operators for Parseval frames. Theorem 16. Let P, Q be projections of rank N on H M and let {e i } M be the coordinate basis of H M. Further, assume that there is a Parseval frame {ϕ i } M for HN with analysis operator T satisfying T ϕ i = P e i for all i = 1,,..., M. If P e i Qe i < ɛ, then there is a Parseval frame {ψ i } M for H M with analysis operator T 1 satisfying T 1 ψ i = Qe i, for all i = 1,,..., M, and

29 1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 9 ϕ i ψ i < ɛ. Moreover, if {Qe i } M is equal norm, then {ψ i} M may be chosen to be equal norm. Proof. By Proposition 7, there are orthonormal bases {a j } M j=1 and {b j} M j=1 for W 1 = T (H N ), W = T 1 (H N ) respectively satisfying a j b j < ɛ. j=1 Let A, B be the N M matrices whose j th columns are a j, b j respectively. Let a ij, b ij be the (i, j) entry of A, B respectively. Finally, let {ϕ i }M, {ψ i }M be the i th rows of A, B respectively. Then we have N ϕ i ψ i = a ij b ij = = j=1 N j=1 a ij b ij N a j b j ɛ. Since the columns of A form an orthonormal basis for W 1, we know that {ϕ i }M is a Parseval frame which is isomorphic to {ϕ i} M. Thus there is a unitary operator U : H M H M with Uϕ i = ϕ i. Now let {ψ i } M = {Uψ i }M. Then ϕ i Uψ i = U(ϕ i) U(ψ i) = ϕ i ψ i ɛ. Finally, if T 1 is the analysis operator for the Parseval frame {ψ i } M, then T 1 is a isometry and since T 1 ψ i = Qe i, for all i = 1,,..., N, if Qe i is equal norm, so is {T 1 ψ i } M and hence so is {ψ i} N. Theorem 17. If g(ɛ, N, M) is the function for the Paulsen Problem and f(ɛ, N, M) is the function for the Projection Problem, then f(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 be a Parseval frame for HN satisfying