Linear Algebra and its Applications

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Linear Algebra and its Applications 434 (011) 1893 1901 Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: www.elsevier.

Zimmer b a Department of Mathematics, Central Michigan University, Mount Pleasant, MI 48859, USA b Department of Mathematics, University of Illinois, Urbana, IL 61801, USA ARTICLE INFO Article

1 Linear Algebra and its Applications 434 (011) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: Robustness and surgery of frames < Sivaram K. Narayan a,, Eileen L. Radzwion a,sarap.rimer a, Rachael L. Tomasino a, Jennifer L. Wolfe a, Andrew M. Zimmer b a Department of Mathematics, Central Michigan University, Mount Pleasant, MI 48859, USA b Department of Mathematics, University of Illinois, Urbana, IL 61801, USA ARTICLE INFO Article history: Received 1 January 010 Accepted 8 November 010 Available online 15 January 011 SubmittedbyS.Fallat AMS classification: 15A03 46C99 65F30 ABSTRACT We characterize frames in R n that are robust to k erasures. The characterization is given in terms of the support of the null space of the synthesis operator of the frame. A necessary and sufficient condition is given for when an (r, k)-surgery on unit-norm tight frames in R are possible. Also a generalization of a known characterization of the existence of tight frames with prescribed norms is given. 010 Elsevier Inc. All rights reserved. Keywords: Frames Tight frames Robustness Erasure Surgery 1. Introduction The focus of this paper is on robustness and surgeries of frames in finite dimensional Hilbert spaces. The study of finite dimensional frames has been motivated by a variety of applications such as signal processing, multiple-antenna wireless systems, and sampling theory. The concept of frames was introduced by Duffin and Schaeffer [5] and popularized by Daubechies [4]. A good introduction to frames in finite dimensional Hilbert spaces can be found in [7]. < Research was supported by NSF-LURE grant Corresponding author. Tel.: ; fax: addresses: sivaram.narayan@cmich.edu (S.K. Narayan), eileen.l.radzwion@gmail.com (E.L. Radzwion), sara.p.rimer@ gmail.com (S.P. Rimer), tomas1r@cmich.edu (R.L. Tomasino), wolfeje@cmich.edu (J.L. Wolfe), zimmen4@illinois.edu (A.M. Zimmer) /$ - see front matter 010 Elsevier Inc. All rights reserved. doi: /j.laa

2 1894 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) A frame in a finite dimensional Hilbert space is a redundant spanning set of vectors. We consider two operations on frames. The erasure consists of removing vectors in a frame. A frame is said to be robust to k erasures if after randomly removing k vectors the resulting set is still a frame. In this paper we give a necessary and sufficient condition for a frame in R n to be robust to k erasures. The condition is stated in terms of the support of the null space of the synthesis operator of the frame. This theorem generalizes a characterization of frames robust to one erasure given by Casazza and Kovačević[]. The second operation that we consider is the removal of r vectors from the frame and adding k vectors to the frame called (r, k)-surgery. We characterize when a (r, k)-surgery is possible on a unit-norm tight frame in R. The result on length surgery" generalizes a characterization of the existence of tight frames with prescribed norms found in [3].. Preliminaries We begin with the definition of a frame. Definition.1. A frame for a Hilbert space H is a sequence of vectors {x i } i I H for which there exist constants 0 < A B < such that for every x H, A x x, x i B x. i I Here A is the greatest lower frame bound and B is the least upper frame bound. A frame is called a tight frame if A = B. Auniform frame is a frame in which all the vectors have equal norm. If all the norms in a uniform frame equals one, the frame is called a unit-norm frame. We focus on frames in finite dimensional Hilbert spaces. Let {x i } i I be a frame in H. The linear map V : H l (I) defined by (Vx) i = x, x i is called the analysis operator. The Hilbert space adjoint V is called the synthesis operator. Theframe operator is given by V V.InR n the analysis operator of a frame {x i } m i=1 is given by the m-by-n matrix x 1 x V =.. x m and the synthesis operator V is given by the n-by-m matrix V = x 1 x... x m and the frame operator is the matrix V V. The following equivalent descriptions of frames in R n are used in this paper. Theorem. [7]. The following statements are equivalent: (1) {x 1, x,...,x m } is a frame in R n. () {x 1, x,...,x m } is a spanning set for R n. (3) The n-by-m matrix V = [ ] x 1 x... x m has rank n.

3 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Robustness Suppose an encoded version Vx of a vector x in R n is transmitted across a communication network. If k of the components of Vx are lost or not delivered the receiver would want to be able to reconstruct Vx using the components that have been received. If V represents the analysis operator of a frame {x i } m i=1 in Rn, what are the frames that allow one to recover the coefficients corresponding to the k frame vectors that are erased" during transmission? Such frames are often called robust to k erasures. Notice that these are exactly the frames that remain frames after any k frame vectors are removed. For more information on this topic the reader may consult [1,3,6,8,9]. We begin with the formal definition of a frame robust to k erasures. Definition 3.1 [3]. Aframe{x i } m i=1 is said to be robust to k erasures if {x i} i I C is still a frame, for any index set I {1,,..., m} with I =k. Now we consider the problem of classifying frames robust to k erasures. The characterization shows that every index set of size m k + 1 is in the support of the null space of the synthesis operator. Definition 3.. The support of a vector x = (x 1, x,...,x m ) in R m, denoted supp(x), is the set of indices {i {1,,...,m} :x i = 0}. Definition 3.3. Let N(A) denote the null space of a matrix A.Thesupport of the null space of A,denoted (A), is the collection of supp(x)asx varies over N(A). That is, (A) ={supp(x) : x N(A)}. Example 3.4. Let A = Then N(A) = span{v 1, v } where v1 T = (1, 0, 1, 1, 1) and vt (A) we consider supp(x) where x = αv 1 + βv, α, β R. = (0, 1, 1, 0, 1). Todetermine Case 1: if α = 0, then the possibilities for supp(x) are or {, 3, 5}. Case : if β = 0, then the possibilities for supp(x) are or {1, 3, 4, 5}. Case 3: if α = 0 and β = 0 then the possibilities for supp(x) are {1,, 3, 4, 5}, {1,, 3, 4}, or {1,, 4, 5}. Hence (A) ={{, 3, 5}, {1, 3, 4, 5}, {1,, 3, 4}, {1,, 4, 5}, {1,, 3, 4, 5}}. Lemma 3.5. If X, Y (A), thenx Y (A). Proof. Suppose x, y N(A) such that X = supp(x) and Y = supp(y).foranyɛ R,letz ɛ = x + ɛy. Clearly z ɛ N(A) and as the ith component of z ɛ is x i + ɛy i, we see that for all but a finite number of ɛ we have supp(z ɛ )= X Y.ThusX Y (A) The following classification of frames robust to one erasure was given by Casazza and Kovačević.

4 1896 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Theorem 3.6 [3]. Let {x i } m i=1 be a frame in Rn. The following are equivalent: (1) {x i } m i=1 is a frame robust to one erasure. () There are scalars c i = 0, for1 i m, such that m c i x i = 0. i=1 Remark 3.7. Statement () is a condition on the support of the null space of the synthesis operator, in particular: {1,,...,m} (V ), where V is the synthesis operator corresponding to the frame {x i } m i=1. This observation motivated the following theorem. Theorem 3.8. Let {x i } m i=1 be a frame in Rn. The following are equivalent: (1) {x i } m i=1 is a frame robust to k erasures. () For all index sets I {1,,...,m} with I =k 1,I C (V ). Here if I {1,...,m} we let I C denote the complement of I, that is I C ={1,...,m}\I. Proof. () (1) Suppose that x j1, x j,...,x jk are erased from {x i } m j=1. We will first show that x j k can be reconstructed from the remaining frame vectors after erasing x j1,...,x jk. By hypothesis, J = {j 1,...,j k 1 } C (V ) so there exists a vector c N(V ) such that supp(c) = J. Therefore 0 = V c = c i x i = c jk x jk + c i x i. i J i J i =j k Since c jk = 0weobtain ( c i c jk x jk = i J i =j k ) x i. In a similar fashion each of x j1,...,x jk 1 can be also be reconstructed from the frame vectors left after erasing x j1,...,x jk. Finally, as {x 1,...,x m } span R n and the span of {x i } i J includes the vectors {x i } i J C we must have that {x i } i J is a spanning set and hence a frame. (1) () Suppose {x i } m i=1 is a frame robust to k erasures. Let I ={j 1, j,...,j k 1 } {1,,...,m}. Let J (V ) be an index set of maximum cardinality disjoint from I. We claim that J = I C. Suppose not, then J I C and we may choose j k J C I C.Since{x i } m i=1 is robust to k erasures, x j k can be reconstructed from frame vectors remaining after the erasure of x j1,...,x jk.thus x jk = c i x i. i I C i =j k and so c i x i = 0 V y = x jk i I C i =j k where y is the vector with components y i = c i when i I C and not equal to j k, y jk = 1, and y i = 0 when i I. Thensupp(y) (V ), j k supp(y), and supp(y) I C. By Lemma 3.5, wealsohave that J supp(y) (V ). This contradicts the assumption that J has maximum cardinality. Hence J = I C (V ).

5 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Example 3.9. Suppose A = Using Theorem. we observe that the columns of A form a frame in R 3.InExample3.4 we found that (A) ={{, 3, 5}, {1, 3, 4, 5}, {1,, 3, 4}, {1,, 4, 5}, {1,, 3, 4, 5}}. Notice that {1} c ={, 3, 4, 5} / (A). Hence from Theorem 3.8, the columns of A do not form a frame robust to erasures. 4. Frame surgery Given a frame, a natural question is whether it can be manipulated in some way to make it a tight frame. Recall that a tight frame is a frame with equal upper and lower bounds. One kind of manipulation of frames in R n is to maintain the lengths of vectors but change the orientation of the vectors. Another kind of manipulation is to add or remove some vectors from a given frame. Definition 4.1 [7]. An (r, k)-surgery on a finite sequence of vectors in R n removes r vectors and adds k vectors to the sequence. Aframe{x i } m i=1 in R can be represented in polar coordinates as x i = a i cos θ i, a i sin θ i where a i is the length of x i and 0 θ i π is the angle between x i and the positive x-axis. Suppose {x i } m i=1 is a tight frame. Then V V = ai where a is the frame bound and I is the identity matrix. Thus V a V = i cos θ i a i cos θ i sin θ i a = a 0 i cos θ i sin θ i a i sin θ i 0 a implies that n a i cos θ i = 0 i=1 a i sin θ i 0 using double-angle formulas. The vector x i = a i cos θ i a i sin θ i is called the diagram vector associated with the frame vector x i. The above discussion from [7] shows the following result. Theorem 4. [7]. Aframe{x i } m i=1 in m R is a tight frame if and only if x i=1 i = 0. We are now ready to prove our main result on frame surgery. Theorem 4.3. Suppose F ={x i } m i=1 is a unit-norm tight frame for R.Thenan(r, k)-surgery on F which leaves a unit-norm tight frame is possible if and only if the sum of the entries in the strict upper triangular part of the Grammian ( x i, x j ) N i,j=1 is bounded by 1 (k N) where N = m rand x 1,..., x N denote the diagram vectors that remain after deleting r frame vectors.

6 1898 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Proof. Let x 1, x,..., x N be the diagram vectors that are present after removing r vectors from the unit-norm tight frame F.LetW denote their vector sum x x N.Ifweweretoaddk unit vectors to F and keep it a unit-norm tight frame then such a (r, k)-surgery on F is possible if and only if W k. Therefore, W = x x N, x x N N N = x i + x i, x j i=1 N 1 = N + i,j=1 i =j N i=1 j=i+1 x i, x j = N + S where S is the sum of the strict upper triangular part of the Grammian [ x i, x j ] N.Hence(r, k)- i,j=1 surgery is possible if and only if N + S k or equivalently S k N. Remark 4.4. If θ ij denotes the angle between vectors x i and x j then from Theorem 4.3 we get N 1 N S = cos (θ ij ) k N. i=1 j=1+i Corollary 4.5. Let {x i } m i=1 be a unit-norm tight frame in R. Suppose k = m where m >. Thenitis always possible to perform a (k + 1, k)-surgery resulting in a new unit unit-norm tight frame. In particular, any unit-norm tight frame consisting of m vectors may be reduced to a unit-norm tight frame consisting of m s vectors where 1 s m. Proof. Suppose m = k + 1. Then removing k + 1 vectors leaves N = m (k + 1) = k vectors. Because cos (θ ij ) 1, using Remark 4.4, weget N 1 N k(k 1) S = cos (θ ij ) i=1 j=i+1 = k N. Hence from Theorem 4.3 it is possible to perform (k + 1, k)-surgery. Suppose m = k. Thenremovingk + 1 vectors leaves N = m (k + 1) = k 1vectors.Again N 1 N k(k 1) S = cos (θ ij ) < k (k 1) i=1 j=i+1 implies from Theorem 4.3 that (k + 1, k)-surgery is possible. It is easy to observe that a repeated application of the two cases in the proof will reduce any unitnorm tight frame of R consisting of m vectors to a unit-norm tight frame of R consisting of m s vectors where 1 s m. Remark 4.6. A construction of the vectors for Corollary 4.5 in the case m = k + 1canbeshownas follows. For θ [0, π),letr θ be the matrix that rotates a vector x in R by θ radians in the counterclockwise direction. That is,

7 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) cos θ sin θ R θ =. sin θ cos θ Let {x i } m i=1 be a unit-norm tight frame for R where m = k + 1. Suppose the vectors x 1,...,x k+1 are removed from {x i } m i=1 and assume that the remaining vectors are not all identical. Define a, b R as a m = x i. b i=k+ To simplify calculations, pick the angle θ such that R θ a a = + b m = R θ x i. b 0 i=k+ Note that {R θ x i } m i=k+ are the diagram vectors of {R θ x i } m.defineφ i=k+ i [0, π)for k + i m, so that R θ x i = cos φ i sin φ i and R θ x i = cos φ i. sin φ i Consider the vectors {y i } k i=1 where ( y i = cos π φ ) k+1+i sin( π φ = sin(φ k+1+i). k+1+i) cos(φ k+1+i ) Then for 1 i k, ỹ i = cos(π φ k+1+i) = cos φ k+1+i. sin(π φ k+1+i ) sin φ k+1+i Consider the sequence F ={R θ x i } m i=k+ {y i} k i=1.clearlyr θ x i and y j are linearly independent when cos φ i = sin φ i.as{x i } m i=k+ are all not identical there must exist vectors R θ x i and y j that are linearly independent. So F spans R and therefore is a frame. Also, m k m R θ x i + ỹ i = cos φ i k + cos φ k+1+i i=k+ i=1 i=k+ sin φ i i=1 sin φ k+1+i a = + b + a + b 0 0 = 0 0 shows that F is a tight frame. When considering tight frames, a natural question arises: when do the norms of vectors in a frame automatically prohibit the frame from being a tight frame? The question is answered in [] for an n-dimensional Hilbert space.

8 1900 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Theorem 4.7 []. There is a tight frame for an n-dimensional Hilbert space H with m vectors having norms x i =a i,i= 1,,...,m if and only if the following inequality is satisfied: max 1 i m{a } 1 m i a. i n i=1 Definition 4.8. Let a 1, a,...,a m denote norms of m vectors in R n.an(r, k)-length surgery on {a i } m i=1 removes r numbers and replaces them with k positive numbers. We have the following generalization of Theorem 4.7. Theorem 4.9. Given a sequence {a i } m i=1 of nonnegative real numbers, it is possible to perform a (0, k)- length surgery resulting in a sequence of nonnegative real numbers that corresponds to norms of vectors in a tight frame for R n if and only if 0 k < nand a = max max {a } 1 m i a 1 i m n. i k i=1 Proof. Let M = m i=1 a i and consider {r i } k i=1 {a i} m i=1.ifa 1 max M, then the fundamental inequality n holds for all k N if r i = 0forall1 i k. Otherwise,a > 1 max n M. With the addition of {r i} k i=1 the left hand side of the fundamental inequality becomes: max{a, max r, 1 r,...,r} k (1) while the right hand side becomes: 1 n (M + r + 1 r + +r). k () As we want Eq. (1) to be as small as possible and Eq. () to be as large as possible, we may assume r := r 1 = = r k.theneq.(1) becomes max{a, max r } while Eq. () becomes 1 n (M + kr ).Define the function h :[0, ) R as (M + kr ) a max if r < a max n h(r) = (M + kr ) r if r a max. n Then a (0, k)-length surgery that results in a sequence that satisfies the fundamental inequality is possible if and only if there exists r 0 such that 0 h(r 0 ). This is equivalent to saying that the global maximum of h is nonnegative. Because h is a piecewise monotone function, the global maximum occurs at either 0, a max,or. Note that h(0) = M n a max, h(a max) = M + ( k ) n n 1 a max, and M + (k n)r lim h(r) = lim = r r n if k < n M/n if k = n if k > n. Summarizing, when k < n we have lim r h(r) <0 and also that h(0) <0. So the desired (0, k)- length surgery is possible if and only if h(a max ) 0, which is equivalent to a max M n k. Acknowledgements This paper presents some of the results obtained during the 008 NSF-LURE summer program at Central Michigan University. Sivaram Narayan was the faculty advisor for this project and Andrew

9 S.K. Narayan et al. / Linear Algebra and its Applications 434 (011) Zimmer was a graduate student mentor. The other co-authors were undergraduate students in the LURE program. The authors thank the referees for helpful suggestions. References [1] B.G. Bodmann, V.I. Paulsen, Frames, graphs and erasures, Linear Algebra Appl. 404 (005) [] P.G. Casazza, M. Fickus, J. Kovačević, M.T. Leon, J.C. Tremain, A physical interpretation of finite frames, Appl. Numer. Harmon. Anal. ( 3) (006) [3] P.G. Casazza, J. Kovačević, Equal-norm tight frames with erasures, Adv. Comput. Math. 39 (006) [4] I. Daubechies, Ten lectures on wavelets, in: CBMS-NSF Regional Conference Series in Applied Mathematics, Philadelphia, SIAM, 199. [5] R.J. Duffin, A.C. Schaeffer, A class of nonharmonic fourier series, Trans. Amer. Math. Soc. 7 (195) [6] V.K. Goyal, J. Kovačević, J.A. Kelner, Quantized frame expansions with erasures, Appl. Comput. Harmon. Anal. 10 (3) (001) [7] D. Han, K. Kornelson, D. Larson, E. Weber, Frames for Undergraduates Volume 40 of Student Mathematical Library, American Mathematical Society, Providence, Rhode Island, 007. [8] R.B. Holmes, V.I. Paulsen, Optimal frames for erasures, Linear Algebra Appl. 377 (004) [9] M. Püschel, J. Kovačević, Real tight frames with maximal robustness to erasures, in: Proc. Data. Compr. Conf., Snowbird, UT (005)..

Linear Algebra and its Applications

Linear Algebra and its Applications 439 (2013) 1330 1339 Contents lists available at SciVerse ScienceDirect Linear Algebra and its Applications journal homepage: www.elsevier.com/locate/laa Maximum robustness