Equiangular Tight Frames with Simplices and with Full Spark in Rd

ISSN 1995-0802, Lobachevskii Journal of Mathematics, 2021, Vol. 42, No. 1, pp. 154–165. c Pleiades Publishing, Ltd., 2021.

S. Ya. Novikov1* (SubmittedbyA.M.Elizarov) 1Samara University, Samara, 443011 Russia Received May 23, 2020; revised August 19, 2020; accepted August 26, 2020

Abstract—An equiangular tight frame (ETF) is an equal norm tight frame with the same sharp angles between the vectors. This work is an attempt to create a brief review with complete proofs and calculations of two directions of research on the equiangular tight frames (ETF): bounds of the spark of the ETF, namely the smallest number of the vectors from ETF that are linearly dependent, and the existence of a regular simplex inside ETF. Tracing these two directions, we go through the case of equality in the Welch estimate, see the connection between RIP (restricted isometry property) and the spark of an ETF,construct a regular simplex using the technique of Naimark complements. We show the connection between equality in the lower estimate of the spark and the presence of a simplex inside ETF. Gram matrix and the matrix of the synthesis operator are calculated for the frame with 10 elements in the space R5. This frame contains a regular simplex and its spark is equal to 4, i.e. is not full. On the other hand you may see an example of the ETF with 6 vectors in R3 without a simplex, but with the spark equal to 4. In such cases the term “full spark” is used, so we have an example of the full spark ETF (FSETF). Finally, there is a proof of the necessary condition for the existence of the FSETF in the space Rd with more than d +1vectors, namely, the number of vectors in such a frame is necessary equal to 2d. DOI: 10.1134/S1995080221010200 Keywords and phrases: equiangular tight frame, simplex, spark

1. INTRODUCTION Let n and d be positive integers with n ≥ d, and let F be either R or C. A ﬁnite frame is a spanning set for a d-dimensional Hilbert space Hd over F that generalizes the notion of a basis by relaxing the { }n need for linear independence. In other words, a family of vectors ϕj j=1 is a frame for a real or complex Hd if there are constants 0

*E-mail: [email protected]

154 EQUIANGULAR TIGHT FRAMES 155 Applying the analysis operator to the synthesis operator yields the Gram matrix Φ∗Φ : Fn → Fn, an ∗ n × n matrix whose (j, j )th entry is (Φ Φ)(j, j )=ϕj ,ϕj . Taking the reverse composition gives the n ∗ H → H ∗ { }n H frame operator ΦΦ : d d, ΦΦ y = ϕj, y ϕj . It is well known that sequences ϕj j=1 in d j=1 { }n H H → H and ϕj j=1 in d have the same Gram matrix if and only if there exists a unitary operator U : d such that Uϕj = ϕj for all j =1,...,n. For the single vector {ϕj } we define the synthesis and analysis operators for each j =1,...,n following [1]: F → H ∗ H → F ∗ ϕj : ,ϕd , jxx = x ϕj; ϕj : d ,ϕj y = ϕj, y . n ∗ ∗ Using this notation we can write the frame operator as ΦΦ = ϕjϕj . j=1 × { }n As such the d n matrix representation of the synthesis operator Φ has the frame elements ϕj j=1 as columns. In particular, if the frame bounds are equal, the frame operator has the form ΦΦ∗ = aI with 1 ∗ ∈ a>0, and so signal reconstruction is rather painless: x = a ΦΦ x, x Hd. In this case the frame is called tight or a-tight. Oftentimes, it is additionally desirable for the frame elements to have equal or unit norms, in these { }n cases the frames are equal norm or unit norm respectively. The frame ϕj j=1 is called equal norm 2 frame if there exists c>0 such that ||ϕj|| = c, j =1,...,n. If the frame is a-tight and equal norm simultaneously, we obtain the following relations between d =dim(H), c and a: n ∗ ∗ 2 da = Tr(aI)=Tr(ΦΦ )=Tr(Φ Φ)= ||ϕj || = nc. j=1 In the case a =1(Parseval or normalized tight frames) we have d = nc, so such equal norm frames always have norms less than 1. The first achievement in the construction of the equal norm tight frame with n vectors in Rd with arbitrary n ≥ d is by A.I. Maltsev [2]. Nowadays it is clear that equal norm tight frames exist for any pair (d, n) with n ≥ d as in the real and also in the complex spaces. The systematic constructions of unit norm tight frames for Rd are based on two interconnected methods, such as Spectral Tetris and Sparsity [3]. The next step in the restriction of the reasonable family of frames is the equiangular frames which are equal norm by definition. { }n ≥ The equal norm frame ϕj j=1 is called an equiangular frame if there exists w 0 such that | |2 ϕj ,ϕj = w for all j = j . Such frames are optimal for various inequalities. For example, let’s look at the (mutual) coherence { }n || ||2 of any sequence of equal norm vectors ϕj j=1 with ϕj = c, j =1,...,nin d-dimensional Hilbert space H over F : d ϕj,ϕ μ := max j . j= j c

In the real case, each vector ϕj spans a line and μ is the cosine of the smallest angle between any pair of these lines. 1/2 H ≥ n−d Welch [4] gives a lower bound on μ in d : μ d(n−1) . This inequality will be proved below. The equality in this estimate draws attention of mathematicians [1, 5]. These papers have become the starting point for the present work. We tried to detail, clarify some of the results and make the presentation self- contained. { }n Fm ≤ Lemma 1. For any vectors ϕj j=1 in (m n) let Φ be the matrix, the jth column of which is ϕj for all j. For any a>0 the following assertions are equivalent:

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 156 NOVIKOV { }n 1) ϕj j=1 forms an a-tight frame for its span; 2) ΦΦ∗Φ = aΦ; 3) (ΦΦ∗)2 = aΦΦ∗; 4) (Φ∗Φ)2 = aΦ∗Φ. { }n ⊂ H ⊂ Fm H ∗ Also, if ϕj j=1 d , then it forms an a-tight frame for d if and only if ΦΦ = aPrHd × H where PrHd denotes the m m matrix of the orthogonal projection on the subspace d. { }n As such, ϕj j=1 forms an ETF for its span if and only if one of the assertions 2)–4) is true { }n || ||2 and ϕj j=1 is equiangular. In this case, with c = ϕj , the dimension d is connected with a, n, c and equiangularity constant w : cn n − d a = ,w= c2 . d d(n − 1) { }n H ⊂ Fm H So, equal norm vectors ϕj j=1 in a subspace d , form an ETF for d if and only if they achieve equality in Welch inequality. { }n Fm H Fm Proof. Fix ϕj j=1 in and a>0, and let d be any d-dimensional subspace of that contains { }n ϕj j=1. 1) ⇔ 2) The synthesis operator y ∈ Fn → Φy ∈ Fm is surjective if and only if its codomain {Φy : y ∈ Fn} = { }n span ϕj j=1. { }n H If we want ϕj j=1 to form an a-tight frame for d, it leads to the equality ∗ ∗ ∈ H ΦΦ = aIHd or ΦΦ x = ax, x d. In this case H { }n { ∈ Fn} d = span ϕj j=1 = Φy : y and we have 2): ΦΦ∗Φy = aΦy, y ∈ Fn. 2) ⇔ 3) H { ∈ Fm} { }n On the other hand, writing d = PrHd x : x , we see that ϕj j=1 forms an a-tight frame for Hd if and only if ∗ ∈ Fm ΦΦ PrHd x = aPrHd x, x . Let’s look at ⎛ ⎞ n n ⎝ ⎠ ∈ Fn PrHd Φy = PrHd y(j)ϕj = y(j)ϕj = Φy, y , j=1 j=1 that is PrHd Φ = Φ. ∗ ∗ ∗ ∗ So we have Φ PrHd =(PrHd Φ) = Φ , and ΦΦ = aPrHd , ∗ ∗ 2 2 2 ∗ ΦΦ ΦΦ = a (PrHd ) = a PrHd = aaPrHd = aΦΦ . 2) ⇒ 3), 2) ⇒ 4) Multiplying 2) by Φ∗ on the right or left gives 3) or 4) respectively. ∗ If either 3) or 4)√ is valid we use the singular value decomposition Φ = UDV , where D is the diagonal matrix with 0 and a on the diagonal. Consequently we obtain DD∗D = aD and then 2). 2 ∗ cn || || − H Now let’s assume that ϕj = c for all j, and let’s look at the Frobenius norm of ΦΦ d Pr d . { }n This quantity is the measure of tightness of ϕj j=1 [5]. Let’s remind that n ∗ ∗ 2 ||A||Fro = Tr(A A)=Tr(AA )= |ai,j| , (1) i,j=1

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 EQUIANGULAR TIGHT FRAMES 157 and Tr(Φ∗Φ)=Tr(ΦΦ∗)=nc. So we have ∗ cn 2 ∗ ∗ cn ∗ cn 2 0 ≤ Tr ΦΦ − PrH = Tr (ΦΦ ΦΦ ) − 2 Tr (ΦΦ PrH )+ Tr (PrH ) d d d d d d n 2 2 | | 2 2 2 (cn) (cn) 2 ϕi,ϕj 2 (cn) = |ϕi,ϕj | − 2 + = c + nc − d d ||ϕi||||ϕj || d i,j=1 i= j |ϕ ,ϕ | 2 c2n(n − d) ≤ c2n(n − 1) max i j − . (2) i= j ||ϕi||||ϕj || d Solving for the coherence here gives the Welch bound: |ϕ ,ϕ | 2 n − d max i j ≥ . i= j ||ϕi||||ϕj || d(n − 1) { }n H | |2 If ϕj j=1 is an ETF for d with ϕi,ϕj = w for all i = j, the two inequalities above become 2 n−d equalities and w = c d(n−1) . Conversely, if Welch inequality is valid as equality, the ﬁnal equality in (2) is 0 implying both { }n 2 inequalities are equalities and so ϕj j=1 is tight and equiangular.

2. THE RIP-PROPERTY AND THE LOWER ESTIMATE FOR THE SPARK { }n ⊂ H Deﬁnition 1. The set of vectors ϕj j=1 d is said to have the restricted isometry property (RIP) for a given integer k and δ ∈ [0, 1) if (1 − δ)||x||2 ≤||Φx||2 ≤ (1 + δ)||x||2 (3) for all k-sparse vectors x ∈ Fn that is for all x ∈ Fn with at most k nonzero entries. The restricted isometry property (RIP) characterizes matrices which are nearly orthonormal, at least when operating on sparse vectors. The concept was introduced by E. Candes and T. Tao and is used to provemanytheoremsintheﬁeld of compressed sensing. { }n ∈ || ∗ − || ≤ Lemma 2 [6]. The set ϕj j=1 (k, δ)-RIP if and only if ΦKΦK I 2 δ for all k-element subsets K of [n]={1,...,n}, where ΦK denotes the synthesis operator of {ϕj }j∈K. Proof. We start with noting that (3) is equivalent to || ||2 −|| ||2 ≤ || ||2 ΦKx 2 x 2 δ x 2 for all K⊆[n],x∈ Fn,xis k-sparse. Then we observe that || ||2 −|| ||2 − ∗ − ΦKx 2 x 2 = ΦKx, ΦKx x, x = (ΦKΦK I) x, x . ∗ For the Hermitian matrix (ΦKΦK − I) we have the Rayleigh–Ritz equality [7]: ∗ ∗ (ΦKΦK − I) x, x ||ΦKΦK − I|| =max , 2 − − || ||2 x k sparse,x=0 x 2 || ∗ − || ≤ 2 and (3) is equivalent to max ΦKΦK I 2 δ. |K|≤k { }n ≥ ∈ { }n ≥ Lemma 3. If ϕj j=1 is (k, δ)-RIP for some k 2 and δ [0, 1), then spark ϕj j=1 k +1. ∗ Proof. Note that since δ<1, from Lemma 2 we obtain that each Gram matrix ΦKΦK is invertible. It { } { }n ≥ 2 seems that any k columns ϕj j∈K are linearly independent, and so spark ϕj j=1 k +1. { }n ≥ ≥ Lemma 4. For equal norm (k, δ)-RIP vectors ϕj j=1 with k 2 the inequality δ μ, where μ { }n is the coherence of ϕj j=1, is valid.

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Proof. Let’s assume that ||ϕj || =1, j =1,...,n.From (k, δ)-RIP we obtain (the resulting matrix has two eigenvalues ±|ϕ ,ϕ |) j j ⎡ ⎤ ∗ 0 ϕj,ϕj ≥ || − || ⎣ ⎦ δ max ΦKΦK I 2 =max =max ϕj,ϕj = μ. |K|=2 j= j j= j ϕ ,ϕ 0 j j 2 2 { }n { }n Theorem 1. Let ϕj j=1 is a set of unit norm vectors and μ ϕj j=1 denotes the coherence of { }n ≤ − this set. If k<1/μ +1, then ϕj j=1 is (k, δ)-RIP with δ (k 1)μ. ∗ Proof. According to Lemma 2 we are to obtain the upper bound to max ||ΦKΦK − I||2, which is |K|=k ∗ equal to the largest eigenvalue of the matrix ΦKΦK − I. It’s important to note that this matrix has zeros on the main diagonal. ∗ As stated in Gershgorin theorem [7] all eigenvalues of the matrix (ΦKΦK − I) lie in the circle with the center at 0 and radius max ϕj ,ϕj ≤ (k − 1)μ. j∈K j∈K,j= j So we obtain the upper bound on the RIP-bound δ ≤ (k − 1)μ. 2 { }n ≥ 1 Corollary 1. We have the inequality spark ϕj j=1 μ +1for any equal norm set of vectors { }n ϕj j=1. { }n ∈ { }n ≥ Proof. If k<1/μ +1, then ϕj j=1 is (k, δ)-RIP for some δ [0, 1) (cf. Th. 1). So spark ϕj j=1 { }n ≥ 2 k +1(cf. Lemma 3). Passing to supremum above k, we obtain spark ϕj j=1 1/μ +1.

3. NAIMARK COMPLEMENTS IN THE BUILDING OF SIMPLICES In this section the well-known technique of Naimark complements will be used to construct some special kinds of ETF, namely, regular simplices. { }s+1 ⊂ H Deﬁnition 2. Let s be a positive integer. The sequence ϕj j=1 d is a regular s-simplex, if s+1 s+1 it is an ETF for the span {ϕj} and dim span {ϕj} = s. j=1 j=1 s+1 s+1 Lemma 5. If {ϕj} is a regular s-simplex, then μ {ϕj } =1/s. j=1 j=1 Proof. According to Lemma 1 the vectors {ϕ }s+1 form an ETF for the span {ϕ }s+1 with j j=1 j j=1 { }s+1 dim span ϕj j=1 = s if and only if s +1− s 1/2 μ {ϕ }s+1 = =1/s. j j=1 s(s +1− 1) 2 { }n H { }n Corollary 2. If ϕj j=1 is an ETF for d and there is a regular s-simplex among ϕj j=1 with s ≤ n, then n − d 1/2 μ {ϕ }n = =1/s. j j=1 d(n − 1) { }n H Proof. As ϕj j=1 is an ETF for d, the “angle” ϕj,ϕj is the same for all pairs j = j . So we have n − d 1/2 1 μ {ϕ }n = = μ {ϕ }s+1 = , j j=1 d(n − 1) jk k=1 s { }n 2 where the vectors ϕj1 ,...,ϕjs+1 form a regular s-simplex contained in ϕj j=1.

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 EQUIANGULAR TIGHT FRAMES 159 { }n { }n { }n Lemma 6. Let ϕj j=1 be an a-tight frame for its span ϕj j=1 and dim span ϕj j=1 = × 1 ∗ d

1 ∗ 2 For P := a Φ Φ we deduce that P = P . Thus, given that P is a Hermitian matrix we get that P is an orthogonal projection matrix on a d-dimensional subspace. As any orthogonal projection matrix it has d eigenvalues 1 and n − d eigenvalues 0, trace(P )=rank(P )=d. From this we obtain that Φ∗Φ = aP has d eigenvalues a and n − d eigenvalues 0, n ∗ 2 ∗ trace(Φ Φ)=ad = ||ϕj || , rank(Φ Φ)=d. j=1 2 { }n Theorem 2. Any a-tight frame for d-dimensional span ϕj j=1 has Naimark complement {ψ }n with the following properties: j j=1 { }n − { }n (i) ψj j=1 is an a-tight frame for the (n d)-dimensional span ψj j=1 . ∗ ∗ (ii) Gram matrix Ψ Ψ = aIFn − Φ Φ. (iii) ΦΨ∗ = 0. − 1 ∗ ∗ 2 Proof. It’s directly verified that the matrix Q := IFn a Φ Φ satisfies Q = Q = Q and its eigenvalues are 1 and 0 with multiplicities n − d and d respectively. It means that the columns of Q, ψ := Qe , j =1,...,n,where {e }n is the orthonormal basis in Fn, defines the span {ψ }n and j j j j=1 j j=1 n dim span {ψj} = n − d. j=1 ∈ { }n If f span ψj j=1 , then Qf = f, ⎛ ⎞ n n n ⎝ ⎠ f = Q Qf,ejej = f,QejQej = f,ψjψj, j=1 j=1 j=1 { }n { }n i.e. the set ψj j=1 forms Parseval (or normalized tight) frame for the span ψj j=1 . √ The vectors ψ = aψ , j =1,...,n form an a-tight frame for the span {ψ }n = j j j j=1 { }n ∗ − ∗ span ψj j=1 and Ψ Ψ = aI Φ Φ. (iii) The Lemma 1 gives that ∗ (ΨΦ∗) (ΨΦ∗)=ΦΨ∗ΨΦ∗ = Φ (aI − Φ∗Φ) Φ∗ = aΦΦ∗ − (ΦΦ∗)2 = 0. The Frobenius norm of the matrix || ∗||2 ∗ ∗ ∗ ΨΦ Fro = trace ((ΨΦ ) (ΨΦ )) = trace0 =0. Consequently, ΨΦ∗ = 0, and its conjugate ΦΨ∗ = 0. 2

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 160 NOVIKOV Corollary 3. If {ϕ }n is an ETF for the span {ϕ }n , then each one of its Naimark j j=1 j j=1 { }n { }n complements ψj j=1 is an ETF for the span ψj j=1 . Proof. In the notation of the Theorem 2 we have that (cf. Lemma 1) (Ψ∗Ψ)2 =(aI − Φ∗Φ)2 = a2I − 2aΦ∗Φ + Φ∗ΦΦ∗Φ = a2I − 2aΦ∗Φ + aΦ∗Φ = a (aI − Φ∗Φ)=aΨ∗Ψ, { }n { }n i.e. (again Lemma 1) ψj j=1 is an a-tight frame for the span ψj j=1 . { }n || ||2 || ||2 −|| ||2 − If ϕj j=1 is an ETF with ϕj = c, j =1 ,...,n,then ψj = a ϕj = a c, j =1,...,n. Besides we have that ψj,ψj = − ϕj,ϕj for all j = j . 2 Corollary 4. A regular s-simplex exists for any positive integer s. { }s+1 | |2 Proof. Let s be a positive integer. Any sequence cj j=1 of scalars ( cj := u>0, j =1,...,s+1) is a (s +1)u-tight equiangular frame for F1. If α ∈ R, then 1 s+1 s+1 α = c αc , (s +1)uα2 = (c α)2; u(s +1) j j j j=1 j=1 if α ∈ C, then 1 s+1 s+1 α = c αc ,u(s +1)|α|2 = |c α|2. u(s +1) j j j j=1 j=1 ∗ Besides, we have that ΦΦ = u(s +1), |c c | = u for all j = j , and ⎛j j ⎞ ⎜ u c1c2 ... c1cs+1⎟ ⎜ ⎟ ⎜ ⎟ ∗ ⎜ c2c1 u ... c2cs+1⎟ Φ Φ = ⎜ ⎟ . ⎜ ⎟ ⎝ ...... ⎠

cs+1c1 cs+1c2 ... u { }s+1 According to the Theorem 2 the Gram matrix for the Naimark complement to the set cj j=1 is the matrix ⎛ ⎞ − − ⎜ su c1c2 ... c1cs+1⎟ ⎜ ⎟ ⎜ − − ⎟ ∗ ∗ ⎜ c2c1 su ... c2cs+1⎟ Ψ Ψ =(s +1)uI − Φ Φ = ⎜ ⎟ . ⎜ ⎟ ⎝ ...... ⎠

−cs+1c1 −cs+1c2 ... su ∗ Any Gram matrix deﬁnes the set of unitary equivalent families of vectors. Looking√ at the matrix Ψ Ψ it is clearly seen that any such family of vectors forms (s +1)u-tight equiangular su-norm frame for { }s+1 s-dimensional span ψj j=1 . { }s+1 || ||2 c(s+1) Conversely, let ϕj j=1 be a regular s-simplex with ϕj = c, j =1,...,s+1. Then a = s and |ϕ ,ϕ | = c/s. So, the Gram matrix for its Naimark complement is j j ⎛ ⎞ c − − ⎜ ϕ1,ϕ2 ... ϕ1,ϕs+1 ⎟ ⎜ s ⎟ ⎜ − c − ⎟ ∗ c(s +1) ∗ ⎜ ϕ2,ϕ1 ... ϕ2,ϕs+1 ⎟ Ψ Ψ = I − Φ Φ = ⎜ s ⎟ , s ⎜ ⎟ ⎝ ...... ⎠ − − c ϕs+1,ϕ1 ϕs+1,ϕ2 ... s

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 EQUIANGULAR TIGHT FRAMES 161 ∗ { }s+1 moreover, all entries of the matrix Ψ Ψ are unimodular. Thus Naimark complement to the ϕj j=1 is unitary equivalent to the set of unimodular scalars. 2

4. SPARK AND SIMPLEX This section is devoted to the connection between equality in the lower estimate of the spark and the presence of a simplex inside ETF. This connection was obtained in [1]. { }n H Theorem 3. If the ETF ϕj j=1 for d contains a regular simplex, then this simplex has s +1 vectors, where 1 d(n − 1) 2 s = (4) n − d i.e. the inverse to Welch bound. { }n H If there are s +1linearly dependent vectors among the ETF ϕj j=1 for d where s is deﬁned by (4), then these s +1vectors form a s-regular simplex. { }n H Proof. Let the ETF ϕj j=1 for d contains a regular simplex for some positive integer s. W. l. o. g. we denote the vectors of the simplex by ϕ1,ϕ2,...,ϕs+1. According to the deﬁnition of a regular s-simplex dim span {ϕ }s+1 = s, i.e. the vectors ϕ ,ϕ ,...,ϕ are linearly dependent. Since j j=1 1 2 s+1 {ϕ }n is equiangular, μ {ϕ }n = μ {ϕ }s+1 . Moreover, both sequences are ETF for H and j j=1 j j=1 j j=1 d { }s+1 span ϕj j=1 respectively, hence we deduce the equality of the Welch bounds for dimensions d and s respectively. In other words, we have that 1 1 n − d 2 (s +1)− s 2 1 = = . d(n − 1) s((s +1)− 1) s

Now suppose that we have s +1linearly dependent vectors ϕ1,ϕ2,...,ϕs+1 with s from (4). Being { }s+1 H ⊂ H linearly dependent, ϕj j=1 is contained in some s-dimensional subspace s d. As a subset of the 1 { }n { }s+1 n−d 2 1 ETF ϕj j=1, ϕj j=1 has coherence d(n−1) = s . The Welch bound for any s +1equal norm vectors s+1 in Hs is 1/s. According to lemma 1 it means that {ϕj} is an ETF for Hs and so is a regular s-simplex. 2 j=1 { }n H Corollary 5. Let ϕj j=1 be an ETF for d. Equality 1 spark {ϕ }n = +1 (5) j j=1 μ { }n is valid if and only if the set ϕj j=1 contains a regular simplex. ⇒ { }n H Proof. ( ) Firstly, as ϕj j=1 is an ETF for d, we have the equality n − d 1/2 μ {ϕ }n = . j j=1 d(n − 1) { }n If (5) is valid, spark ϕj j=1 = s +1, where s is given by (4). By the deﬁnition of the spark it { }n means that ϕj j=1 contains s +1linearly dependent vectors. The second proposition of the Theorem 3 asserts, that these s +1vectors form a s-regular simplex. ⇐ { }n ( ) If the ETF ϕj j=1 contains a regular simplex, then this simplex has s +1linearly dependent vectors with s ftom (4). So we obtain the inequality spark {ϕ }n ≤ s +1. On the other hand, we j j=1 { }n ≥ 2 see (cf. Cor. 1) that spark ϕj j=1 s +1.

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 162 NOVIKOV { }n Corollary 6. If there are simplices within the ETF ϕj j=1, then the simplices are the smallest linearly dependent subsets of {ϕ }n . Moreover, any simplex is a full spark system for its span. j j=1 { }n Proof. The number of vectors in an arbitrary simplex exactly coincides with the spark ϕj j=1 { }n ≥ H (cf. Cor. 5). Recall that the set of vectors ϕj j=1, n d +1, in d is a full spark system if { }n { }n spark ϕj j=1 = d +1, i.e. any d vectors from the set ϕj j=1 are linearly independent. In the s+1 particular case when an ETF is a simplex {ϕj } , any s vectors within it are linearly independent. 2 j=1 Example 1. Let’s take a closer look at the example from [1]. We see the 6 × 10 matrix ⎡ ⎤ ⎢ 1111111111⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1111−1 −1 −1 −1 −1 −1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 −1 −1 −1111−1 −1 −1⎥ Φ=⎢ ⎥ . (6) ⎢ ⎥ ⎢−11−1 −11−1 −11 1−1⎥ ⎢ ⎥ ⎢ ⎥ ⎣−1 −11−1 −11−11−11⎦ −1 −1 −11−1 −11−11 1 Considering the matrix Φ as the matrix of the synthesis operator, we calculate the matrices of the frame operator and the Gram matrix: ⎡ ⎤ ⎢ 10 −2 −2 −2 −2 −2⎥ ⎢ ⎥ ⎢ ⎥ ⎢−210−2 −2 −2 −2⎥ ⎢ ⎥ ⎢− − − − − ⎥ ∗ ⎢ 2 210 2 2 2⎥ ΦΦ = ⎢ ⎥ , ⎢ ⎥ ⎢−2 −2 −210−2 −2⎥ ⎢ ⎥ ⎢ ⎥ ⎣−2 −2 −2 −210−2⎦ −2 −2 −2 −2 −210 ⎡ ⎤ ⎢ 6222222−2 −2 −2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 26222−2 −22 2−2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2262−22−22−22⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2226−2 −22−22 2⎥ ⎢ ⎥ ⎢ − − − ⎥ ∗ ⎢ 22 2 262222 2⎥ Φ Φ=⎢ ⎥ . ⎢ ⎥ ⎢ 2 −22−22622−22⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 −2 −22226−22 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢−22 2−22 2−2622⎥ ⎢ ⎥ ⎢ ⎥ ⎣−22−22 2−22262⎦ −2 −22 2−222226 ∗ 2 ∗ { }10 Moreover, calculations show that (ΦΦ ) = 12ΦΦ . It means (cf. Lemma 1) ) that the columns ϕj j=1 of the matrix Φ are 12-tight frame for its span. Besides this frame is equiangular with c =6, μ =1/3. cn 6·10 The dimension of the span may be deduced from the d = a = 12 =5. Really we see, that any ϕj lies

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t 6 in 5-dimensional orthogonal complement to the vector(1, 1, 1, 1, 1, 1) in R . So we have the ETF with { }10 10 vectors for its 5-dimensional span, and μ ϕj j=1 =1/3. In this case, since the number s determined by the equation (4) is equal to 3, to solve the question of the existence of a simplex within the system, it suﬃces to ﬁnd 4 linearly dependent vectors { }10 among ϕj j=1(cf. Th. 3) and all such vectors were found [1]. Moreover, according to corollary 6, { }10 spark ϕj j=1 =4.

5. PHASE RETRIEVAL AND FULL SPARK ETF IN Rd In this section you may ﬁnd the interconnection between sparks of Naimark complement frames. Then we make an outlook on Phase Retrieval problem, and one of the main results from this area will help us to reconstruct the proof of the necessary condition for the existence full spark ETF in Rd. A close assertion to the following was obtained in [8]. { }n Rd { }n Lemma 7. If ϕj j=1 is an a-tight full spark frame in , then its Naimark complement ψj j=1 is an a-tight full spark frame in Rn−d. { }n Rn−d Proof. Firstlynotethat ψj j=1 is an a-tight frame in (cf. Th. 2). So we have to prove only, { }n that Ψ= ψj j=1 is a full spark frame. As usual we can look at Ψ as a matrix of the synthesis operator. We suppose the contrary, Ψ is not a full spark frame, it means that Ψ has a collection of n − d linearly dependent columns, i.e. there exists a nontrivial sequence {αj}j∈K such that αjψj = 0, j∈K and |K| = n − d. By the deﬁnition of the synthesis operator, we have that ψj =Ψδj, where δj is an identity basis element in Rn,so ⎛ ⎞ ∗ ∗ ⎝ ⎠ ∗ ∗ 0 =Ψ 0 =Ψ αjψj = αjΨ Ψδj = αj (aIRn − Φ Φ) δj. j∈K j∈K j∈K Now we denote by ∗ x := a αjδj = αjΦ Φδj. (7) j∈K j∈K

∗ d Let’s note that x = 0 since {αj}j∈K is nontrivial, x ∈ Range(Φ Φ) = R . Furthermore, whenever i ∈K, we have from (7) that ⎛ ⎞ ∗ ∗ ∗ ⎝ ⎠ 2 x, Φ Φδi = Φ Φx,δi = a Φ Φ αjδj ,δi = a αjδj,δi =0, j∈K j∈K

∗ ∗ d and so x ⊥ span{Φ Φδi}i∈Kc . Thus, the containment span{Φ Φδi}i∈Kc ⊆ R is proper, ∗ ∗ d = Rank(Φ Φ) > Rank (Φ ΦKc )=Rank (ΦKc ) , i.e. d × d-submatrix ΦKc is rank-deficient, and therefore Φ is not full spark. So we obtain a contradiction with the assumption about Φ. 2 To make the proof of the final theorem of our survey independent, we have to observe briefly the problem that appears in the literature under various names. Phase Retrieval (Phase Recovery, Phaseless Reconstruction, or briefly PR) is called the problem of signal reconstruction by absolute values of linear measurements. Let Rˆd be the quotient space Rd/ ± 1, the set of equivalence classes x ∼ y with respect to x = ±y. Consider the following nonlinear map ˆd n 2 β : R → R :(β(ˆx))j = |x, ϕj | ,j =1,...,n.

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 164 NOVIKOV { }n A set of vectors Φ= ϕj j=1 admits a Phase Retrieval Property (PRP) if the mapping 2 2 t βΦ = |ϕ1,x| ,...,|ϕn,x| {| |2}n is injective. Numbers ϕj ,x j=1 are called signal intensity measurements. { }n Rd Deﬁnition 3. A set of vectors Φ= ϕj j=1 in satisﬁes the complement property (CP) if for d every subset T ⊆{1,...,n} either {ϕj}j∈T or {ϕj}j∈T c spans R . More expressive combination is the alternative completeness. The properties (CP) and (PRP) are known to be equivalent in Rd.

Lemma 8.[9] The mapping βΦ is injective iﬀ Φ satisﬁes the complement property.

Proof. (⇒) Suppose that Φ ∈ (CP). Hence there is T ⊆{1,...,n} such that {ϕj }j∈T , and d d {ϕj }j∈T c are not complete in R . Let non zero vectors u, v ∈ R be such that u, ϕj =0for all j ∈ T, c and v, ϕn =0for all j ∈ T . For any j we have 2 2 2 2 2 |u ± v, ϕj | = |u, ϕj | ± 2u, ϕj v, ϕj + |v, ϕj | = |u, ϕj | + |v, ϕj | .

2 2 So |u + v, ϕj| = |u − v, ϕj | for any j, and βΦ(u + v)=βΦ(u − v). As u and v are non zero, we have that u + v = ±(u − v). So the mapping βΦ is not injective. d (⇐) Suppose that βΦ is not injective. It means that there exist vectors x, y ∈ R such that x = ±y and βΦ(x)=βΦ(y). We denote by T := {j : x, ϕj = −y,ϕj}. We have that x + y,ϕj =0for any c j ∈ T, and x − y,ϕj =0for any j ∈ T . By assumption we have that x = ±y, so x + y =0 and d x − y =0 . So we have that {ϕj}j∈T and {ϕj }j∈T c are not complete in R . 2 { }n Rd ≤ ≤ − Theorem 4 [1]. If ϕj j=1 is a full spark ETF for and 2 d n 2, then n =2d. { }n Rd ≤ ≤ − Proof. Let ϕj j=1 is a full spark ETF for where 2 d n 2, and assume that n =2d. Naimark { }n Rn−d − complement ψj j=1 is a full spark ETF for (Lemma 7). The numbers n and n d are connected by the Gerson bound n ≤ (n − d)(n − d +1)/2 [10, 11] or (n − d)2 ≥ n + d. The last inequality is valid only for n ≥ 2d whenever d ≥ 3. So we may assume w. l. o. g. that n ≥ 2d +1. The equiangularity gives equalities

|ϕj,ϕ2d| = |ϕj,ϕ2d+1| ,j=1, 2,...,2d − 1. ± { }n Rd ≥ Let’snotethatϕ2d = ϕ2d+1, since ϕj j=1 is an ETF for with d 2, and so we have that |ϕ2d,ϕ2d+1|

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 42 No. 1 2021 EQUIANGULAR TIGHT FRAMES 165 Example 2. ⎛ ⎞ ⎜0011ν −ν⎟ 1 ⎜ ⎟ Φ=√ ⎜11ν −ν 00⎟ , 1+ν2 ⎝ ⎠ ν −ν 0011 √ √ 1+ 5 5 ν = 2 . Φ is the ETF with μ(Φ) = 5 . We may apply Corollary 1 to obtain 1 √ spark(Φ) ≥ +1= 5+1, μ and so spark(Φ) = 4, i.e. Φ is a full spark ETF in R3. This frame does not contain any regular simplex, 1/2 √ 3·5 because s = 6−3 = 5 (cf. Theorem 3). In the space R4 the ETF with 8 vectors is impossible according to the following theorem from [12, th. 17]STDH07: Theorem 5. If there exists a real d × 2d ETF, then d is odd and (2d − 1) is the sum of two squares. { }10 In the Example 1 we may see the ETF with 10 vectors ϕj j=1 in its 5-dimensional span. But this { }10 frame is not full spark, we’ve seen above that spark ϕj j=1 =4. It looks like the question of the existence the FSETF with 10 vectors in R5 is still open.

FUNDING The work of the author was carried out as part of the implementation development programs of the Scientiﬁc and Educational Mathematical Center Volga Federal District, Agreement no. 075-02-2020- 1488/1.

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