Spanning and independence properties of frame partitions

We answer a number of open problems in frame theory concerning the decomposition of frames into linearly independent and/or spanning sets. We prove that in finite dimensional Hilbert spaces, Parseval frames with norms bounded away from 1 can be decomposed into a number of sets whose complements are spanning, where the number of these sets only depends on the norm bound. We also prove, assuming the Kadison-Singer conjecture is true, that this holds for infinite dimensional Hilbert spaces. Further, we prove a stronger result for Parseval frames whose norms are uniformly small, which shows that in addition to the spanning property, the sets can be chosen to be independent, and the complement of each set to contain a number of disjoint, spanning sets.


Introduction
A family of vectors {f i } i∈I is a frame for a Hilbert space H if there are constants 0 < A ≤ B < ∞ satisfying The theory of frames in Hilbert spaces has applications covering a broad spectrum of problems in pure mathematics, applied mathematics and engineering [4]. Many fundamental questions in frame theory involve determining the extent to which frames can be decomposed into subsets which to some extent resemble bases. It is known that these problems are generally difficult to resolve. For example, work of the second author shows that decomposing frames into subsets which are Riesz basic sequences is equivalent to an important open problem in analysis -the 1959 Kadison-Singer Problem [4].
In this note we will answer a number of open problems concerning the decomposition of frames into linearly independent and/or spanning sets. The solutions to these problems, for finite dimensional Hilbert spaces, requires some non-trivial variations of the Rado-Horn Theorem, which is itself rather delicate. We prove that some infinite dimensional analogues of these results would have a positive answer if the Kadison-Singer Problem has a positive answer. In particular, the fact that the number R appearing in Theorem 3.2 can be chosen independent of the dimension of the underlying Hilbert space is implied by the assumption that the Kadison-Singer Problem has a positive The first author was supported by NSF Grant DMS-0807399, the second author by NSF DMS-0704216, the third author by NSF DMS-0600191 and the fourth author by NSF DMS-0354957. 1 answer. Indeed, one of the motivations that led to the study of these consequences of Kadison-Singer was, initially, a quest for a negative answer to the Kadison-Singer Problem. However, our results verify that these consequences of a positive answer to the Kadison-Singer Problem are, in fact, true.
In Section 2, we derive a result about Parseval frames with norms bounded away from 1 that is a consequence of the assumption that the Kadison-Singer Problem has a positive answer. We will see that these results involve questions about spanning sets. In Section 3, we then show that the finite dimensional versions of these results are uniformly true, that is, with constants that do not depend on the dimension, without the need to assume that the Kadison-Singer Problem has a positive answer. The results of Section 3 are in some sense refinements of the Rado-Horn theory and rely strongly on earlier refinements of Rado-Horn obtained by the second and fourth authors together with Kutyniok. Section 4 is a further development which investigates the consequences of having Parseval frames with uniformly small norms.

Kadison-Singer and Spanning Properties for Frame Partitions
In this section, we begin with a few observations about Parseval frames and prove a result about spanning sets for Parseval frames whose norms are bounded away from 1, assuming that the Kadison-Singer Problem has a positive answer.
Given a family of vectors F = {f i } i∈S in a Hilbert space H, where S is some index set, and a subset B ⊆ S, we write F B = {f i } i∈B and let H B denote the closed linear span of F B . Recall that if {f i } i∈S is a Parseval frame for H and P is the orthogonal projection onto some closed subspace, then {P f i } i∈S is a Parseval frame for P (H). Also, recall that {f i } i∈S is a Parseval frame for H if and only if the Gram matrix G = ( f j , f i ) i,j∈S is the matrix of a projection operator on ℓ 2 (S).
We begin with a few useful observations. Proposition 2.1. Let {f i } i∈S be a Parseval frame for H, let P be an orthogonal projection onto a closed subspace of H and let I denote the identity operator on H. Then Proof. The equality G = R + Q is immediate from R = ( P f j , f i ) and Q = ( (I − P )f j , f i ) for each i, j ∈ S, and from the linearity of the inner product in the first entry. The fact that G, Q and R are matrices of projections follows from the fact that the vectors {f i } i∈S form a Parseval frame. For the final statement, note that P = I if and only if Q = 0. But since Q is a projection, Q = 0 if and only if 1 is not an eigenvalue.
Given a subset B ⊆ S, we let D B = (d i,j ) i,j∈S denote the bounded operator on ℓ 2 (S) whose matrix is the diagonal matrix with d i,i = 1 when i ∈ B and d i,j = 0 when i ∈ B c or j ∈ B c , the complement of the set B.
Proof. Let P denote the projection onto H B and apply Proposition 2.1. Since for j ∈ B, f j ∈ H B , we have that when j ∈ B then f j , f i = P f j , P f i . More generally, the matrices G and R are equal in any entry (i, j) provided that i ∈ B or j ∈ B. Thus, the matrix Q must be 0 in any such entry. Hence we obtain the operator inequalities Now, if 1 is not an eigenvalue of D B c GD B c , then these inequalities imply that 1 is not an eigenvalue of Q. Invoking the preceding proposition, we get P = I, and so Conversely, assume that 1 is an eigenvalue of Proof. Note that the set {P e j : j ∈ S} is a Parseval frame for P (H). Hence, the span of {P e j : j ∈ B} is dense in P (H) if and only if the matrix Q = ( P e j , P e i ) i,j∈Bc does not have 1 as an eigenvalue. But since I ℓ 2 (B c ) − Q = ( (I − P )e j , (I − P )e i ) i,j∈B c , Q not having eigenvalue 1 is equivalent to the latter matrix having a trivial kernel. For our next result, we will be assuming that the Anderson Paving Problem has an affirmative answer. The Anderson Paving Problem and the Kadison-Singer Problem are known to be equivalent [1]. There are several equivalent versions of Anderson's Paving Problem. The particular version that we shall use asserts the following: For each 0 < s < 1, there exists an r depending only on s, such that if H = (h i,j ) ∈ B(ℓ 2 (N)) is any operator with h i,i = 0, for every i, then there exists a partition of N into r disjoint sets A 1 ∪ · · · ∪ A r = N, with D A k HD A k ≤ s H , for k = 1, ..., r.
Theorem 2.5. Let 0 < δ < 1. If the Anderson Paving Problem has a positive answer, then there exists an r depending on δ, such that whenever {f n } n∈N is a Parseval frame for a Hilbert space H, with f n 2 ≤ 1 − δ for all n ∈ N, then there exists a partition of N into r disjoint sets, A 1 ∪ · · · ∪ A r = N, such that H A c k = H, for k = 1, ..., r.
has 0 diagonal and H ≤ 1. Set s = δ/2 in the statement of Anderson's Paving and let r be the corresponding integer. Then we may pick disjoint sets, and it follows that 1 can not be an eigenvalue. From the preceding proposition, it follows that The study of Parseval frames with norms bounded away from 1 is in some sense complementary to other results relating Parseval frames and the Kadison-Singer Problem, e.g. [2], since most other work relating these problems focuses on Parseval frames whose norms are bounded away from 0 rather than 1 and focuses on linear independence rather than spanning. However, having Parseval frames that are norm-bounded away from 1 has the added advantage that when one projects onto a subspace, then the projections of these vectors is a Parseval frame for the subspace that is bounded away from 1 by the same bound. In contrast, a Parseval frame with norms that are bounded away from 0 might no longer have norms bounded away from 0 when one projects it onto a subspace.

Spanning properties for partitions of Parseval frames with norms bounded away from one
The previous section illustrates that Anderson's paving would provide a partition of certain norm-bounded Parseval frames {f i } i∈I into a number of sets with specific spanning properties. Now we show that the existence of such a number, r, can be obtained independently of the assumption of Anderson Paving, if the Hilbert space is finite dimensional. Moreover, our choice of r depends only on the norm bound 1 − δ, and not the dimension of the space, and an explicit formula for r as a function of δ is provided.
Recall that a matroid is a finite set X together with a collection of subsets of X, I, which satisfy three properties: (1) ∅ ∈ I (2) if I 1 ∈ I and I 2 ⊂ I 1 , then I 1 ∈ I, and (3) if I 1 , I 2 ∈ I and |I 1 | < |I 2 |, then there exists x ∈ I 2 \ I 1 such that I 1 ∪ {x} ∈ I. We will say that elements of I are independent. We also recall that the rank of a set E ⊂ X is defined to be the cardinality of a maximal independent set contained in E. Now, given a set of vectors {f j : j ∈ J} which spans H N , we say J ∈ J if {f j : j ∈ J} spans H N . It is straightforward to verify that (J, J ) forms a matroid. Indeed, the first two properties are immediate and the third property reduces after taking complements to the fact that if {f j : j ∈ E 1 } and {f j : j ∈ E 2 } both span H N , and |E 1 | > |E 2 |, then there exists x ∈ E 1 \ E 2 such that {f j : j ∈ E 1 , j = x} spans.
We note here that for a natural number n, rank(E) ≥ n if and only if there is a set F ⊂ E such that |F | = n and {f j : j ∈ F } spans H N .
Finally, we recall the Rado-Horn Theorem [6,7] in the context of matroids.
Theorem 3.1. [5] Let (X, I) be a matroid, and let R be a positive integer. A set J ⊂ X can be partitioned into R independent sets if and only if for every subset E ⊂ J, Since a Parseval frame must span, we have that (J, J ) is a matroid. By the Rado-Horn Theorem, it suffices to show (3.1) for each subset of J. Let E ⊂ J. Define S = span{f j : j ∈ E}, and let P be the orthogonal projection onto S ⊥ . Since the orthogonal projection of a Parseval frame is again a Parseval frame, we have that {P f j : j ∈ J} is a Parseval frame for S ⊥ . Moreover, we have Let M be the largest integer smaller than or equal to |E|(1−δ). Since dim S ⊥ ≤ M, we have that there exists a set as desired.
We note that it is not possible, in general, to get the partition in Theorem 3.2 to have the property that the {f j } j∈A i are linearly independent. Again, the problem is that we do not have a lower bound on the norms of the frame vectors and so there can be an arbitrarily large number of them. That is, there can be too many frame vectors to be able to partition them into R linearly independent sets. However, we will see that it is possible to achieve a partition in which all sets but one are linearly independent and spanning, if the norms of the vectors are uniformly small.

Spanning and linear independence properties for Parseval frames with uniformly small norms
In this section we obtain a strengthening of the preceding section with the help of a generalization of the Rado-Horn Theorem due to Casazza, Kutyniok and Speegle [3]. (1) There exists a partition {I j } M j=1 of I so that for each j, {f i } i∈I j is linearly independent.
(2) For all J ⊂ I, Moreover, in the case that the above conditions fail, there exists a partition {I j } M j=1 of I and a subspace S of X such that the following three conditions hold.
(a) For all 1 ≤ j ≤ M, S = span {f i : i ∈ I j , and f i ∈ S}.
In particular, for each 1 ≤ j ≤ M, {f i : i ∈ I j , f i / ∈ S} is linearly independent.
We also need a slight generalization of a result of Casazza and Tremain [4]. can be partitioned into r + 1 linearly independent sets. If k = 0, {f i } rN i=1 can be partitioned into r linearly independent spanning sets.
is an equal norm Parseval frame, we have That is, , for all i = 1, 2, . . . , N.
We will verify that the assumption of the Rado-Horn Theorem holds for r + 1. Choose J ⊂ {1, 2, . . . , rN + k}. Let P be the orthogonal projection of H N onto span {f i } i∈J . Since {P f i } i∈J is a Parseval frame for its span we have That is, The result now follows by the Rado-Horn Theorem and the fact that in the case k = 0, we have partitioned an rN element set into r linearly independent sets in an N-dimensional Hilbert space H N , and hence, each must contain exactly N elements and so it must be a spanning set.
We now want to strengthen Proposition 4.2 to show that we can actually partition our family of vectors into a linearly independent set and r linearly independent spanning sets. Proof. For all j = 1, 2, . . . , r, the fact that {f i } i∈I j are linearly independent implies that the dimension of the span of {f i } i∈I j = |I j |. Also, the fact that {f i } i∈A j span H N implies |A j | ≥ N. Now, we have Hence,  In particular, the number of frame vectors in a unit norm frame with lower frame bound A is greater than or equal to⌊A⌋N.
Proof. We replace {f i } i∈I by { 1 √ r f i } i∈I so that our frame has lower frame bound greater than or equal to 1 and Assume the frame operator for {f i } i∈I has eigenvectors {e j } N j=1 with respective eigenvalues λ 1 ≥ λ 2 ≥ . . . λ N ≥ 1. We proceed by induction on N.
it follows that |{i ∈ I : f i = 0}| ≥ r and so we have a partition into r spanning sets.
Assume the inductive hypothesis holds for H N and consider H N +1 .
We check two cases: Case I: Suppose there exists a partition {I j } r j=1 of I so that {f i } i∈I j is linearly independent for all j = 1, 2, . . . , r.
Case II: Our family cannot be partitioned into r linearly independent (spanning) sets.
In this case, let {I j } r j=1 and a subspace ∅ = S ⊂ H N +1 be given by Theorem 4.1. If S = H N +1 , we are done. Otherwise, let P be the orthogonal projection onto the subspace S. Let Theorem 4.1 (c) implies that {f i } i∈I ′ j is linearly independent for all j = 1, 2, . . . , r. To see this, note that the non-zero elements of {(I − P )f i } i∈I are {(I − P )f i } i∈I ′ . Fix 1 ≤ j ≤ r and assume there are scalars This implies by Theorem 4.1 (c): i∈I ′ j α i f i ∈ S, and so α i = 0, for all i ∈ I ′ j .
Applying the induction hypothesis, we can find a partition {A j } r j=1 of I ′ with span {(I − P )f i } i∈A j = (I − P )(H N +1 ), for all j = 1, 2, . . . , r.

Now, we can apply Lemma 4.3 together with the partition
Note that we cannot expect to get any linear independence in Proposition 4.4 because our vectors can have arbitrarily small norms and hence there can be an arbitrarily large number of them. However, we can remove appropriate vectors from the last r − 1-sets until they are linearly independent and spanning. Putting the removed vectors into the first set, we get a partition into a spanning set and r − 1 linearly independent spanning sets.
Corollary 4.5. Let {f i } i∈I be a Parseval frame for H N and r a natural number so that f i 2 ≤ 1 r for every i ∈ I. Then there is a partition {I j } r j=1 of I so that span {f i : i ∈ I j } = H N , for all j = 1, 2, . . . , r.
In the following we answer a question concerning the partition of equal norm Parseval frames into spanning sets which continues Proposition 4.2, and had been left open in [4].
We will be considering partitions which maximize dimensions in a very particular way. Proof. Assuming the hypothesis of the lemma, if f k = l∈Ip,l =k α l f l , then removing f k from I p keeps dim span(F Ip ) constant. By property (MD), moving f k into another I j , j = p cannot increase dim span(F I j ), and the result follows.
If there are linear dependent sets in a partition having property (MD) then we can move suitable vectors from one set to another. The following definition will be used to help us keep track of which vectors are being moved.  1 , b 1 ), . . . , (a n , b n )}, where a i ∈ I and b i ∈ {1, . . . , M}, such that • (a 1 , b 1 ) is a chain of length one, • for 2 ≤ i ≤ n, a i ∈ I b i and f a i = αf a i−1 + j∈I b i ,j =a i α j f j for some α = 0, and • a i = a k for i = k. A chain of length n starting with a 1 ∈ L ⊂ I and ending at a n ∈ I is a chain of minimal length starting in L and ending at a n if every chain starting in L and ending at a n has length greater than or equal to n.
We recall the following lemma. If {(a 1 , b 1 ), . . . , (a n , b n )} is a chain of minimal length starting in L and ending at a n , then for each 1 ≤ i ≤ n, {(a 1 , b 1 ), . . . , (a i , b i )} is a chain of minimal length starting in L and ending at a i .
Proof. By induction it suffices to show that {(a 1 , b 1 ), . . . , (a n−1 , b n−1 )} is a chain of minimal length. Suppose, for the sake of contradiction, that there did exist a chain {(u 1 , v 1 ), . . . , (u k , v k )} such that u k = a n−1 and k < n − 1. Since {(a 1 , b 1 ), . . . , (a n , b n )} is a chain, f an = αf a n−1 + j∈I bn ,j =an α j f j for some α = 0. Therefore, either {(u 1 , v 1 ), . . . , (u k , v k ), (a n , b n )} is a chain with length k + 1 < n or a n = u i for some i ≤ k, either of which contradicts the minimality of n. Proof. The set of partitions of I into M sets has a partial ordering with respect to which two partitions {1, 2, . . . , M}. I is a finite set, so there are maximal elements. By definition, these partitions have the property (MD).
Assume that there is a partition with property (MD) which contains more than one set for which the associated vectors are linearly dependent, say I 1 and I 2 . We can then successively remove indices from I 2 and place them into I 1 if the associated vectors are linear combinations of others remaining in the set indexed by I 2 . After finitely many such moves, F I 2 is linearly independent. Moreover, by Lemma 4.7, the span of F I 1 and F I 2 retain their dimensions, which means the maximality is preserved.
If F I 2 , . . . , F I M are linearly independent, L = {i ∈ I 1 : f i = j∈I 1 ,j =i α j f j }, and {(a 1 , b 1 ), . . . , (a n , b n )} is a chain of minimal length starting in L, it follows that for each 1 ≤ i < n, b i = b i+1 . In this case, we can track the changes in the partition as vectors are moved among the sets in a straightforward manner.
Definition 4.11. If F I 2 , . . . , F I M are linearly independent, then proceeding by induction, we can define and for 2 ≤ i ≤ n, Let L be as above and assume that {(a 1 , b 1 ), . . . , (a n , b n )} is a minimal chain starting in L. For each 1 ≤ i ≤ n, f a i can then be written as the sum Proof. For the case i = 1, note that a 1 ∈ L implies that f a 1 = j∈L,j =a 1 α j f j for some choice of α j . By Lemma 4.9 none of these j ∈ L can be in {a p : 1 ≤ p ≤ n} since this would not be a chain of minimal length starting in L. Recalling that b i = 1, the claim is proven for i = 1.
Proceeding by induction, let i ∈ {1, . . . , n} and we assume (4.2) is true for 1 ≤ k < i. We will show that it is also true for i. Note that where we have used in the last two lines that , suppose for the sake of contradiction that there is a j ∈ I b i ∩ U i b i such that α j = 0 and j = a p for some p > i. Then {(a 1 , b 1 ), . . . , (a i−1 , b i−1 ), (a p , b i )} is a chain starting in L, which contradicts the minimality of the chain {(a 1 , b 1 ), . . . , (a n , b n )}. So, using the induction hypothesis on each term in the last sum in (4.4) and combining terms, one obtains with an apporpriate choice ofα j 's.  {(a 1 , b 1 ), . . . , (a n , b n )} is a chain of minimal length starting in L and ending at a n , then f an ∈ span(F Lm ) for all 1 ≤ m ≤ M.
Proof. We show that, if {(a 1 , b 1 ), . . . , (a n , b n )} is a chain of minimal length starting in L and ending at a n , then f an ∈ span(F Lm ) for each 1 ≤ m ≤ M.
By Lemma 4.12 and the fact that ). In particular, the partition {U i k : 1 ≤ k ≤ M} satisfies property (MD). By property (MD), Lemma 4.12, and Lemma 4.7, f an ∈ span(F U n m ) for each 1 ≤ m ≤ M. Therefore, for m = b n , there exist α By definition of a chain, for each a p such that b p+1 = m and 1 ≤ p < n − 1, for some choice of α (p) j and some α (p) = 0. Fix j 0 such that α (0) j 0 = 0 in (4.5). We show that j 0 ∈ L m , which finishes the proof of the lemma. Clearly, if j 0 ∈ {a 1 , . . . , a n }, then we are done, so we assume that j 0 ∈ {a 1 , . . . , a n }.
The purpose of this is to prove the following: Theorem 4.14. Let {f i } i∈I be a finite collection of vectors in a finite dimensional vector space X. Assume (1) {f i } i∈I can be partitioned into r + 1-linearly independent sets, and (2) {f i } i∈I can be partitioned into a set and r linearly independent spanning sets. Then there is a partition {I i } r+1 i=1 so that {f j } j∈I i is a linearly independent spanning set for all i = 2, 3, . . . , r + 1 and {f i } i∈I 1 is a linearly independent set.
Proof. We choose the partition {I i } r+1 i=1 of I that maximizes dim span {f j } j∈I 1 taken over all partitions so that the last r sets span X. If {J i } r+1 i=1 is a partition of I such that for all 1 ≤ i ≤ r + 1, dim span {f j } j∈J i ≥ dim span {f j } j∈I i , then dim span {f j } j∈I i = dim span {f j } j∈J i for all i = 2, . . . , r+1 since dim span {f j } j∈I i = dim X, and dim span{f j } j∈I 1 = dim span{f j } j∈J 1 by construction. This means, the chosen partition has the property (MD) and the properties asserted by Lemma 4.13. Suppose that this does not partition F I into linearly independent sets, i.e. F I 1 is not linearly independent. As in Lemma 4.13, let L = {i ∈ I 1 : f i = j∈I 1 ,j =i α j f j } be the index set of the "linearly dependent vectors" in I 1 , L 0 = {i ∈ I : there is a chain starting in L ending at i}, and L j = L 0 ∩ I j , 1 ≤ j ≤ r + 1.
Let S = span(F L 0 ). By Lemma 4.13, S = span(F L j ) for all 1 ≤ j ≤ r + 1. Moreover, for 1 ≤ j ≤ r + 1, i ∈ L j implies that i ∈ I j and f i ∈ S. Therefore, S ⊂ span{f i : i ∈ L j } ⊂ span{f i : i ∈ I j , f i ∈ S} = S.
Let J = {i ∈ I : f i ∈ S}. By construction, L ⊂ J. Let d = dim(S) and see that, by the preceding portion of this proof, dim span(F J ) = d. Moreover, |J| = |L 1 | + · · · + |L M | = |L 1 | + rd > d(r + 1), because L 1 is linearly dependent, since it contains L by virtue of chains of length one. Therefore, for J = {i ∈ I : f i ∈ S}, |J| dim span(F J ) > r + 1. This is in contradiction with assumption (1), which implies by the Rado Horn theorem that |J|/d ≤ r + 1.
Combining Propositions 4.2 and 4.4 with Theorem 4.14 we conclude: Corollary 4.15. Let {f i } i∈I be an equal norm Parseval frame for H N with |I| = rN +k with 0 ≤ k < N. Then there is a partition {I i } r+1 i=1 of I so that for i ∈ {2, . . . , r + 1}, {f j } j∈I i is a linearly independent spanning set and {f j } j∈I 1 is linearly independent.
If r ≥ 2 then this result implies that each set of frame vectors has a complement which is spanning, which was already obtained in Section 3. The insight of this last corollary is that with the lower norm bound implicit in the equal-norm Parseval property, the partition can be chosen to consist of linearly independent sets. Moreover, the complement of each set can then be partitioned into at least r − 1 spanning sets.