On the existence of complex Hadamard submatrices of the Fourier matrices


 We use a theorem of Lam and Leung to prove that a submatrix of a Fourier matrix cannot be Hadamard for particular cases when the dimension of the submatrix does not divide the dimension of the Fourier matrix. We also make some observations on the trace-spectrum relationship of dephased Hadamard matrices of low dimension.

where P i and D i are permutation and diagonal matrices, respectively.
It is easy to see that every Hadamard matrix is equivalent to a dephased Hadamard matrix. Hadamard matrices up to and including dimension N = 5 have been classified up to equivalence [1], but even for dimension N = 6, no complete classification exists. Dutkay, Haussermann, and Weber proved that through dimension N = 5, two dephased Hadamard matrices with the identical traces will be spectrally equivalent [2]. However, a counterexample was known for the 12×12 case. At the end of this paper we make a note that counterexamples in fact exist in the N = 6 case. A subclass of the dephased N × N Hadamard matrices are the Fourier matrices F N , given by The Fourier matrices and their Hadamard submatrices are a subclass of the Butson-type Hadamard matrices. The Butson class BH(n, q) consists of n × n Hadamard matrices whose entries are qth roots of unity. Thus, F N ∈ BH (N, N). We refer the reader to [3] for further information on Butson-type Hadamard matrices. is Hadamard. However, in this paper we will stick to the ν = 1 case, as it affords the following equivalence between Hadamard triples and submatrices of Fourier matrices: The following lemma is straightforward and will allow us to prove this proposition: )m, and k 1 and k 2 are distinct since k ′ 1 and k ′ 2 are not congruent modulo m. Likewise, for every j ′ ∈ J ′ there exists a unique j ∈ J and integer ℓ j such that j ′ = j + a + ℓ j m. Then by virtue of the fact that H * H = nIn, where n = |J| = |K|, we have that: Hadamard triples were used by Jorgensen and Pedersen in [4] to demonstrate that, for a measure µ induced by an iterated function system (IFS) with parameters N and B, if an L can be found so that (N, B, L) is a Hadamard triple, then the exponential functions For example, because the quaternary Cantor measure µ 4 is induced by an IFS with parameters N = 4, and is complete in L 2 (µ 4 ), and therefore an orthonormal basis of L 2 (µ 4 ). The existence of an orthogonal basis of complex exponential functions means that µ 4 is a spectral measure with spectrum Λ. On the other hand, the famous ternary Cantor measure µ 3 is induced by N = 3 and B = {0, 2}, but there is no L so that (3, {0, 2} , L) is a Hadamard triple. The nonexistence of the Hadamard triple does not a priori imply that µ 3 is not spectral, but Jorgensen and Pedersen went on to show that it is, in fact, not spectral. Moreover, in [5] Dutkay and Jorgensen showed that the Fuglede Conjecture, which is still unresolved in both directions in dimensions 1 and 2, is in dimension 1 in the spectral-tile direction equivalent to a Universal Tiling Conjecture (UTC). The UTC conjectures that if equally-sized sets of integers all share a common spectrum (with respect to the counting measure), then there exists a single translation set T by which each of them will tile Z. For a finite set A ⊆ Z, one can test whether a set Λ, |Λ| = |A|, is a spectrum for A by checking is Hadamard, completeness not being at issue in finite dimensions. If Λ ⊂ Q, then H will be a submatrix of a Fourier matrix, and for an integer N such that NΛ ⊂ Z, (N, NΛ, A) will be (after suitable translations and modulations) a Hadamard triple. Hadamard triples are therefore instrumental in the harmonic analysis of measures, especially those induced by iterated function systems and those that are fractal, and we would like to understand better when they can and cannot be found. In this paper we will ask the following question: For a given N and n, does there exist a Hadamard triple (N, B, L) such that |B| = |L| = n? In light of Proposition 1, and because of the potentially broader interest, we choose to reframe the question in terms of Hadamard submatrices of Fourier matrices as follows: For a given N and n, does there exist an n × n Hadamard submatrix of F N ?

The existence of Hadamard submatrices of the Fourier matrix
The following observation is immediate: If n | m, then there exists an n × n Hadamard submatrix of Fm. This raises the question as to whether the condition n | m is not just sufficient, but also necessary. The theorem of Lam and Leung [6] dealing with zero-sums of roots of unity, which we will utilize below, seems to suggest that counterexamples could possibly be found when m is sufficiently composite. On the other hand, Hadamard matrices have additional orthogonality structure that may preclude this possibility. The remainder of this section eliminates a wide range of subcases of the n m case, but by no means all of them. We first check off our list the following basic impossibility: Our main theorem, which eliminates many instances of the n m case, is as follows: where J, K ⊆ {0, . . . , m − 1}, |J| = |K| = n. Now, each of the n elements of J is a representative of only one congruence class of integers modulo ℓ, but there are ℓ such congruence classes, and since ℓ < n, there are more elements of J than there are congruence classes modulo ℓ. It follows by the Pigeonhole Principle that there must exist two members of J, say r 1 and r 2 , that are congruent modulo ℓ. Without loss of generality, we may suppose r 1 < r 2 , with r 2 = r 1 + bℓ, where b is a positive integer. By the Lam-Leung Theorem, there do not exist m ℓ th roots of unity x 1 , . . . , xn such that x 1 + · · · + xn = 0. However, because the rows r 1 and r 2 of H must be orthogonal, This is a contradiction. Hence, no such H exists. Proof. Take ℓ = 1 in Theorem 1. Since the prime divisors p 1 , . . . , ps are all larger than n, certainly n cannot be of the form n = k 1 p 1 + · · · + ks p 2 for nonnegative integers k j . While Theorem 1 applies to a wide variety of cases, it does not apply to every case where n m. For example, consider the case n = 6, m = 27 = 3 3 . The only choices for ℓ are ℓ = 1 and ℓ = 3, but since 3 is a factor of m/ℓ either way and 6 = 2 · 3, Theorem 1 cannot eliminate this case. A corollary of the Lam-Leung Theorem will allow us to achieve another result that will eliminate this case, too.
Here ζp = e 2πi/p . By a "minimal vanishing sum," it is meant that no proper subsum of the terms also sums to zero.
We are prepared to prove the following result:

Theorem 2. Suppose m = p a where p is an odd prime, and a ∈ N. Then Fm does not have a Hadamard submatrix of size 2p.
Proof. The a = 1 case is trivial, so let a ≥ 2. Assume, for the sake of contradiction, that Fm does have a Hadamard matrix of size 2p. Then by Lemma 1, it has a 2p × 2p submatrix H that uses the first row and first column of Fm, so that the first row and first column of H consist entirely of 1's. Let us say Hr,s = e 2πijr ks /m , where J = {j 1 , j 2 , . . . , j 2p } ⊂ Zm, K = {k 1 , k 2 , . . . , k 2p } ⊂ Zm, and j 1 = k 1 = 0. In light of the orthogonality of H, this means that the entries of any row besides the first row must sum to zero, and likewise the entries of any column besides the first column must sum to zero. Since such sums are zero sums of p a th roots of unity, by the Lam-Leung Corollary, each of them must allow a partition into minimal subsums, each being a sum of a rotation of the pth roots of unity. In fact, since the number 1 is in each row and column, every row and column of H (except the first row and column) has at least one set of the unrotated pth roots of unity. Assume that there exist two columns of H besides the first column, say s, s ′ ∈ {2, 3, . . . , 2p}, that both consist only of the unrotated pth roots of unity, so that each partitions into two sets of the pth roots. So there is some r ≠ 1 such that Hr,s = 1. For a positive integer x, let α(x) denote the number of times p divides into x. It follows that α(jr) + α(ks) ≥ a, because 1 = Hr,s = e 2πijr ks /p a . Now, the rth row of H cannot contain any more 1's, so for a different row r ′ / ∈ {1, r}, we have H r ′ ,s ′ = 1. It follows that α(j r ′ ) + α(k s ′ ) ≥ a. Then α(jr) + α(j r ′ ) + α(js) + α(j s ′ ) ≥ 2a. It follows that either α(jr) + α(k s ′ ) ≥ a or α(j r ′ ) + α(ks) ≥ a. Therefore, either m | j r ′ ks or m | jr k s ′ . So either H r ′ ,s = 1 or H r,s ′ = 1, and either way, this is a contradiction to the fact that H cannot contain more than two 1's in any column besides the first column. Therefore, H cannot have two columns in addition to the first column that consist entirely of unrotated pth roots of unity. A similar argument holds for the rows of H.
Therefore, there must exist some row of H, say R, that contains a copy of the pth roots rotated by (an mth root) ω, where ω p ≠ 1. Let C ⊂ {2, 3, . . . , 2p} be the set of columns of H for which c ∈ C means H R,c is a pth root rotated by ω. We claim that every entry in columns C C := {1, 2, . . . , 2p} \ C is an unrotated pth root of unity. This is obviously true for the first row. Let r ∈ {2, 3, . . . , 2p} be any other row, and let s ∈ C C . If Hr,c is a rotated pth root for all c ∈ C, then the fact that row r must contain the unrotated pth roots leaves no choice but for Hr,s to be an unrotated pth root. Otherwise, there must exist some column c ∈ C such that Hr,c is an unrotated pth root. It follows that α(jr) + α(kc) ≥ a − 1. In addition, since H R,s is an unrotated pth root, we have α(j R ) + α(ks) ≥ a − 1. Thus α(jr) + α kc + α(j R ) + α(ks) ≥ 2(a − 1), and it follows that either α(jr) + α(ks) ≥ a − 1 or α(j R ) + α(kc) ≥ a − 1. The latter cannot be the case, or else H R,c would be a pth root. Therefore, the former is the case, which implies Hr,s is a pth root.
Thus every entry in the columns C C consists only of pth roots. However, this is a contradiction, since there cannot be two columns besides the first with all entries pth roots. Therefore, there is no 2p × 2p Hadamard submatrix of Fm. There are still many cases unhandled by either Theorem 1 or Theorem 2. For example, let m = 45 = 5 · 3 2 and n = 6. Neither theorem applies to this case, but n m.

Trace and spectra of the 6 × 6 Fourier matrix and cyclic 6-roots matrix
While [2] showed that trace equivalence implies spectral equivalence (and, of course, vice versa) for dephased Hadamard matrices up to size 5×5, a counterexample was evidently known in dimension 12×12. We observe in this section that 6 × 6 is the first size for which trace-spectral equivalence for dephased Hadamard matrices does not hold. Consider the following dephased Hadamard matrices: First, consider the 6 × 6 Fourier matrix, which may be written as: 6 . The so-called Cyclic 6-Roots Matrix is defined in [1] as: Also, if we take the 2 × 2 Fourier matrix and the 3 × 3 Fourier matrix where ω = e 2πi 3 , we may form the following two Kronecker products of F 2 and F 3 to obtain two more 6 × 6 dephased Hadamard matrices: where ω = e Recall that Specht's Theorem states that two matrices A and B are unitarily equivalent if and only if tr(W(A, A * )) = tr(W(B, B * )) for any word W. It is known, however, that it is sufficient for this trace equality to hold only for words up to a certain finite length d that depends on the size n of the matrices. A theorem of Pappacena [7] shows that it is sufficient to check words of length no greater than n √︁ Of course, the upper bound on d in Proposition 4 is rather crude, for it is the same bound as that which works for matrices in general when invoking Specht's Theorem. As noted before, in the case of dephased Hadamard matrices of sizes n ≤ 5, [2] showed that it is sufficient to check only d = 1. Our counterexample for n = 6 shows that checking higher values of d is necessary. An open question is to what the bound can be reduced in the special case of dephased Hadamard matrices. We note that tr(C 2 6 ) = 36 ≠ 12 = tr(F 2 6 ), but we do not know whether checking d ≤ 2 is sufficient for all 6 × 6 dephased Hadamard matrices.