On Ryser's Conjecture

TL;DR

This paper establishes new sufficient conditions that confirm Ryser's conjecture in certain cases, advancing understanding of this longstanding combinatorial problem.

Contribution

It introduces novel sufficient conditions that ensure Ryser's conjecture holds, contributing new theoretical insights.

Findings

New sufficient conditions for Ryser's conjecture

Conditions applicable to specific combinatorial structures

Enhanced understanding of conjecture's validity scope

Abstract

We provide new sufficient conditions under which Ryser's conjecture holds.

Equations

$$R(x):=h_1+h_{2}x+\cdots +h_n{x}^{n-1}.$$

$$E_1:=\mathrm{circm}(h_1,h_3,h_5,\dots ,h_{n-1}),E_2:=\mathrm{circm}(h_2,h_4,h_6,\dots ,h_n).$$

$$|c_1+\cdots +c_{\ell }|=\ell ,$$

$$c_1+c_2+\cdots +c_{2k}=0.$$

$$v=(d_1,d_2,d_3,\dots ,d_{2k-2},d_{2k-1},d_{2k}),$$

$$w=(d_{2k},d_1,d_2,\dots ,d_{2k-3},d_{2k-2},d_{2k-1}),$$

$$s:=d_1+d_2+\cdots +d_{2k-1}+d_{2k}\ne 0.$$

$$u=(d_{2k-1},d_{2k},d_1,\dots ,d_{2k-4},d_{2k-3},d_{2k-2}).$$

$$d_j=a{d}_{j+2}+b{d}_{j+1}.$$

$$s=as+bs.$$

$$a+b=1,$$

$$h_k=a{h}_{\ell }+b{h}_m.$$

$$h_k=h_{\ell }=h_m,$$

$$(a,b)\in \{(0,1),(1,0)\}.$$

$$h_k=(a+b)h_{\ell }=h_{\ell }.$$

$$h_{\ell }=1,h_m=-1.$$

$$h_k=a-b.$$

$$c_1=\dots =c_m.$$

$$s\in \{0,2k,-2k\}.$$

$$|R(s)|=\sqrt{n}.$$

$$\sum _{\begin{array}{c}j\ne 1\\ 1\le j\le n/2\end{array}}⟨R_1,R_j⟩+\sum _{\begin{array}{c}j\ne 1\\ 1\le j\le n/2\end{array}}⟨S_1,S_j⟩=\sum _{\begin{array}{c}j\ne 1\\ 1\le j\le n\end{array}}⟨T_1,T_j⟩=0,$$

$$⟨R_1,R_j⟩+⟨S_1,S_j⟩=⟨T_1,T_{2j-1}⟩,$$

$${\lambda }_1:=h_1+h_3+\cdots +h_{n-1},{\lambda }_2:=h_2+h_4+\cdots +h_n,$$

$$h_1+h_3+\cdots +h_{n-1}=0,|h_2+h_4+\cdots +h_n|=\sqrt{n}.$$

$$a:=\sum _{\begin{array}{c}j\ne 1\\ 1\le j\le n/2\end{array}}⟨R_1,R_j⟩,b:=\sum _{\begin{array}{c}j\ne 1\\ 1\le j\le n/2\end{array}}⟨S_1,S_j⟩.$$

$$⟨R_1,R_1⟩=\frac{n}{2}=⟨S_1,S_1⟩.$$

$$a+\frac{n}{2}=⟨R_1,\sum _{1\le j\le n/2}{R}_j⟩=⟨R_1,{\lambda }_1{v}_0⟩={\lambda }_1^2,$$

$$b+\frac{n}{2}=⟨S_1,\sum _{1\le j\le n/2}{S}_j⟩=⟨S_1,{\lambda }_2{v}_0⟩={\lambda }_2^2.$$

$${\lambda }_1^2+{\lambda }_2^2=n.$$

$${\lambda }_1+{\lambda }_2=h_1+h_2+\cdots +h_n\in \{\sqrt{n},-\sqrt{n}\}.$$

Full text

1. Introduction

Let A be a square matrix of order $n$ with complex coefficients. We denote by $A^{*}$ the conjugate transpose of the matrix $A$. For any pair of rows $R_{i},R_{j}$ of a matrix $A$ with real coefficients, we denote by $\langle R_{i},R_{j}\rangle$ the usual scalar product. A circulant matrix $A:=\operatorname{circm}(a_{1},\ldots,a_{n})$ of order $n$ is a matrix of order $n$ whose first row is $[a_{1},\ldots,a_{n}]$ and in which each row after the first is obtained by a cyclic shift of its predecessor by one position. For example, the second row $R_{2}$ of $A$ is $[a_{n},a_{1},\ldots,a_{n-1}]$. With $\pi:=\operatorname{circm}(0,1,0,\ldots,0)$, one has that $R_{2}$ is the first row of $\pi A$.

A Hadamard matrix $H$ of order $n$ is a matrix of order $n$ with entries in $\{-1,1\}$ such that $K:=\frac{H}{\sqrt{n}}$ is orthogonal. A circulant Hadamard matrix of order $n$ is a circulant matrix that is also Hadamard. Besides the two trivial matrices of order $1$, $H_{1}:=\operatorname{circm}(1)$ and $H_{2}:=-H_{1}$, the remaining $8$ known circulant Hadamard matrices are $H_{3}:=\operatorname{circm}(1,-1,-1,-1),H_{4}:=-H_{3},H_{5}:=\operatorname{circm}(-1,1,-1,-1),H_{6}:=-H_{5},$ $H_{7}:=\operatorname{circm}(-1,-1,1,-1),H_{8}:=-H_{7},H_{9}:=\operatorname{circm}(-1,-1,-1,1),H_{10}:=-H_{9}.$

If $H=\operatorname{circm}(h_{1},\ldots,h_{n})$ is a circulant Hadamard matrix of order $n$, then its representer polynomial is

[TABLE]

No other circulant Hadamard matrices are known. Ryser conjectured in 1963 (see [19], [1, p. 97]) that no such matrices exist when $n>4$. Previous work on the conjecture include [2, 3, 6, 7, 8, 9, 11, 13, 15, 16, 21].

Let $H=\operatorname{circm}(h_{1},\ldots,h_{n})$ be a circulant Hadamard matrix of order $n\geq 4$. Define

[TABLE]

When $n=4$, both $E_{1}$ and $E_{2}$ have rank equal to $1$ (as their determinant is 0, see the matrices $H_{3},\ldots,H_{10}$).

Two other interesting properties of $E_{1}$ and $E_{2}$ when $n=4$ are the following:

Remark 1.1*.*

Let $K_{1}:=\frac{E_{1}+E_{2}}{2}$ and $K_{2}:=\frac{E_{1}-E_{2}}{2}$. Then:

• (a)

$K_{1}$ and $K_{2}$ are projections.

• (b)

$K_{1}$ and $K_{2}$, reduced modulo $2$, are symmetric orthogonal matrices.

Property (b) in Remark 1.1 is significant because of a result of McWilliams [12, Corollary 1.8] (see also Lemma 2.3) which states that for $n>2$, the orthogonal group $\mathbb{O}(n,\mathbb{F}_{2})$ consists only of the identity matrix. In other words, $n=4$ is the only case in which (b) holds.

When a circulant Hadamard matrix of order $n\geq 4$ is such that both $E_{1}$ and $E_{2}$ are non-singular, it follows from [9] and our Lemma 2.12 (see Section 2) that $n$ must equal $4$. The rank-$1$ property that we consider in the present paper is the first natural case to study in order to make progress on the open case when at least one of $E_{1},E_{2}$ is singular.

In this paper we prove two results. Essentially, the first shows that for circulant Hadamard matrices of order $n$, the rank-$1$ property and properties (a) and (b) in Remark 1.1 hold for $n=4$ and do not hold when $n>4$. Our second main result, which follows from a simple characterization of circulant matrices of rank $\leq 2$ together with the Plotkin bound for codewords, proves the more general statement that there is no circulant Hadamard matrix of order $>4$ in which either one of the submatrices $E_{1},E_{2}$ has rank $1$ while the other has arbitrary rank, or both have rank $2$.

More precisely, in this paper we prove the following two theorems.

Theorem 1.2**.**

Let $H=\operatorname{circm}(h_{1},\ldots,h_{n})$ be a circulant Hadamard matrix of order $n\geq 4$. Let $E_{1}:=\operatorname{circm}(h_{1},h_{3},h_{5},\ldots,h_{n-1})$ and $E_{2}:=\operatorname{circm}(h_{2},h_{4},h_{6},\ldots,h_{n})$. We denote by $R_{j}$, respectively $S_{j}$, the rows of $E_{1}$ and $E_{2}$, for $j=1,\ldots,n/2$. Let $G_{1}:=E_{1}E_{1}^{*}$ and $G_{2}:=E_{2}E_{2}^{*}$ be the Gram matrices of $E_{1}$ and $E_{2}$. Assume that one of the following statements holds:

• (a)

$\operatorname{rank}(E_{1})=1=\operatorname{rank}(E_{2})$.

• (b)

All entries of $G_{1}$ and $G_{2}$ have the same absolute value.

• (c)

$R_{1}$* and $R_{2}$ are linearly dependent over $\mathbb{Q}$, and $S_{1}$ and $S_{2}$ are linearly dependent over $\mathbb{Q}$.*

• (d)

$|\langle R_{1},R_{2}\rangle|=|\langle R_{1},R_{1}\rangle|$, and $|\langle S_{1},S_{2}\rangle|=|\langle S_{1},S_{1}\rangle|$.

Then $n=4$.

Remark 1.3*.*

All statements (a), (b), (c), and (d) of the theorem hold when $n=4$. Moreover, for any entry $e$ of $G_{1}$ or $G_{2}$, we have $|e|=2=n/2$.

Remark 1.4*.*

Our results also cover the recent generalization of the circulant Hadamard conjecture [5] by Goyeneche and Turek, since a generalized circulant Hadamard matrix $H(d):=\operatorname{circm}(d,h_{2},\ldots,h_{n})$ satisfying our hypotheses necessarily has $|d|=1$.

Our second main result is as follows:

Theorem 1.5**.**

With the notation of Theorem 1.2, there is no circulant Hadamard matrix $H$ of order $n>4$ in the following two cases:

• (a)

One of $E_{1},E_{2}$ has rank $1$.

• (b)

Both $E_{1}$ and $E_{2}$ have rank $2$.

The needed preliminaries for the proofs are given in Section 2. The proof of Theorem 1.2 is presented in Section 3. The proof of Theorem 1.5 is presented in Section 4.

2. Preliminaries

Our first lemma is well known and easy to prove.

Lemma 2.1**.**

Let $A$ be a real matrix. Then $\operatorname{rank}(A^{*}A)=\operatorname{rank}(AA^{*})=\operatorname{rank}(A)=\operatorname{rank}(A^{*})$.

The following lemma ([1]) is useful.

Lemma 2.2**.**

Let $C:=\operatorname{circm}(c_{1},\ldots,c_{k})$ be a circulant matrix of order $k$. Let $v:=[1,\ldots,1]\in\mathbb{R}^{k}$. Then $v$ is an eigenvector of $C$ with associated eigenvalue $\lambda:=c_{1}+\cdots+c_{k}$.

The following is a result of MacWilliams [12]. It was already used in [10, Theorem 1].

Lemma 2.3**.**

The only circulant, symmetric, and orthogonal matrix over the binary field $\mathbb{F}_{2}$ of order $n>2$ is the identity matrix $I_{n}$.

The next lemma (see [20, Lemma 8.6]) is frequently used in the theory of group representations. Here it is useful for the proof of part (d) of Theorem 1.2.

Lemma 2.4**.**

Let $c_{1},\ldots,c_{\ell}$ be $\ell$ complex numbers of absolute value $1$. If

[TABLE]

then $c_{1}=\cdots=c_{\ell}$.

The following lemma gives some properties of certain circulant matrices of rank $1$.

Lemma 2.5**.**

Let $C:=\operatorname{circm}(c_{1},\ldots,c_{2k})$ be a circulant matrix of order $2k$ and rank $1$ such that $c_{j}^{2}=1$ for all $j=1,\ldots,2k$. Then for all $s=1,\ldots,k$ we have $c_{2s-1}=a$ and $c_{2s}=b$, with $\{a,b\}\subseteq\{-1,1\}$. Moreover, if $a\neq b$ then

[TABLE]

Proof.

Since $\operatorname{rank}(C)=1$, the second row $R_{2}$ of $C$ is a multiple of the first row $R_{1}$. As each entry of $C$ is $1$ or $-1$, there are two cases to consider: (a) $R_{1}=R_{2}$ and (b) $R_{1}=-R_{2}$.

In case (a), comparing the first entries of $R_{1}$ and $R_{2}$ gives $c_{2k}=c_{1}$, and comparing the others shows $c_{1}=c_{2}=c_{3}=\cdots=c_{2k-1}$. Thus all entries of $R_{1}$ are equal, so $a=b\in\{-1,1\}$.

In case (b), a similar argument gives the stated result.

If $a\neq b$, then $\{a,b\}=\{-1,1\}$, so $c_{1}+c_{2}+\cdots+c_{2k}=a+b=0$. ∎

The following lemma gives some properties of certain circulant matrices of rank $2$.

Lemma 2.6**.**

Let $D:=\operatorname{circm}(d_{1},\ldots,d_{2k})$ be a circulant matrix of order $2k\geq 6$ and rank $2$ such that $d_{j}^{2}=1$ for all $j=1,\ldots,2k$. Assume that the first two rows of $D$,

[TABLE]

and

[TABLE]

are $\mathbb{Q}$-linearly independent. Let $u=av+bw$ be any row of $D$, where $a,b\in\mathbb{Q}$. Assume that

[TABLE]

Then $a+b=1$.

Proof.

Without loss of generality, we may assume that $u$ is the third row of $D$, so that

[TABLE]

Since $u=av+bw$, with indices taken modulo $2k$, we have for all $j=1,\ldots,2k$:

[TABLE]

Summing over all $j$ in (2) gives

[TABLE]

As $s\neq 0$, equation (3) implies that $a+b=1$, as claimed. ∎

Lemma 2.7**.**

Let $h_{k},h_{\ell},h_{m}\in\{-1,1\}$. Assume that for some rational numbers $a,b\in\mathbb{Q}$ satisfying

[TABLE]

we have

[TABLE]

Then either

[TABLE]

or

[TABLE]

Proof.

Assume first that $ab\neq 0$. It suffices to prove that $h_{\ell}=h_{m}$, since if this holds then (5) and (4) give

[TABLE]

To prove $h_{\ell}=h_{m}$, assume without loss of generality that

[TABLE]

From (5) we obtain

[TABLE]

If $h_{k}=1$, then together with (4) we find $b=0$, a contradiction to $ab\neq 0$. Similarly, if $h_{k}=-1$ we find $a=0$, again a contradiction. Thus $h_{\ell}=h_{m}$.

If $ab=0$, then (4) implies (7). ∎

Lemma 2.8**.**

Let $C=\operatorname{circm}(c_{1},\ldots,c_{m})$ be a circulant matrix such that two consecutive rows are equal. Then

[TABLE]

Proof.

Without loss of generality, assume that the first two rows of $C$ are equal. Then we have $c_{j}=c_{j-1}$ for all $j$, with indices taken modulo $m$. This proves the claim. ∎

Lemma 2.9**.**

Let $C=\operatorname{circm}(c_{1},\ldots,c_{2k})$ be a circulant matrix with entries in $\{-1,1\}$, of order $2k\geq 6$, such that the first row of $C$ is equal to the third. Let $s=c_{1}+\cdots+c_{2k}$. Then

[TABLE]

Proof.

Comparing the first and third rows of $C$ gives $c_{j}=c_{j-2}$ for all $j$ (indices modulo $2k$). Hence, all odd-indexed $c_{j}$ are equal to some $c\in\{-1,1\}$, and all even-indexed $c_{j}$ are equal to some $d\in\{-1,1\}$. Therefore, $s=k(c+d)$, which yields $s\in\{0,2k,-2k\}$. This proves the claim. ∎

The following two Lemmas are well known; see, e.g., [4, p. 1193], [14, p. 234], [21, pp. 329–330].

Lemma 2.10**.**

Let $H$ be a regular Hadamard matrix of order $n\geq 4$. Then $n=4h^{2}$ for some positive integer $h$. Moreover, if $H$ is circulant then $h$ is odd. Furthermore, either $H$ or $-H$ is $2h$-regular (the other being $(-2h)$-regular). In the $2h$-regular case, each row has $2h^{2}+h$ positive entries and $2h^{2}-h$ negative entries; in the $(-2h)$-regular case, each row has $2h^{2}-h$ positive entries and $2h^{2}+h$ negative entries.

Lemma 2.11**.**

Let $H$ be a circulant Hadamard matrix of order $n$, let $w=\exp(2\pi i/n)$, and let $R(x)$ be its representer polynomial. Then:

• (a)

All eigenvalues $R(s)$ of $H$, where $s\in\{1,w,w^{2},\ldots,w^{n-1}\}$, satisfy

[TABLE]

• (b)

The vector $v:=[1,\ldots,1]\in\mathbb{R}^{n}$ is an eigenvector of $H$ with associated eigenvalue $\lambda=\sqrt{n}$.

The following lemma is important for the proof of the theorems.

Lemma 2.12**.**

Let $H:=\operatorname{circm}(h_{1},\ldots,h_{n})$ be a circulant Hadamard matrix, let $E_{1}:=\operatorname{circm}(h_{1},h_{3},h_{5},\ldots,h_{n-1})$, and $E_{2}:=\operatorname{circm}(h_{2},h_{4},h_{6},\ldots,h_{n})$. Then

[TABLE]

where $R_{j}$, $S_{j}$, and $T_{j}$ denote the $j$-th row of $E_{1}$, $E_{2}$, and $H$, respectively.

Proof.

The result follows from the equality

[TABLE]

since two distinct rows of $H$ are orthogonal. ∎

Lemma 2.13**.**

Let $H:=\operatorname{circm}(h_{1},\ldots,h_{n})$ be a circulant Hadamard matrix, let $E_{1}:=\operatorname{circm}(h_{1},h_{3},h_{5},\ldots,h_{n-1})$, and $E_{2}:=\operatorname{circm}(h_{2},h_{4},h_{6},\ldots,h_{n})$. Let

[TABLE]

be the real eigenvalues of $E_{1}$ and $E_{2}$ associated with the eigenvector $v_{0}:=[1,\ldots,1]\in\mathbb{R}^{n/2}$. Then $\lambda_{1}\lambda_{2}=0$, so that we may assume, without loss of generality (possibly replacing $H$ by $\pi H$), that

[TABLE]

Proof.

With the same notation as in Lemma 2.12, put

[TABLE]

Since $h_{j}^{2}=1$ for all $j$, we have

[TABLE]

Thus,

[TABLE]

and

[TABLE]

By Lemma 2.12 we have $a+b=0$. Adding the two equations above gives

[TABLE]

Moreover, Lemma 2.11 implies

[TABLE]

These two relations together yield $\lambda_{1}\lambda_{2}=0$, as claimed. ∎

In order to prove our second theorem we will recall a special case of the Plotkin bound (see [17], [18]) as well as a simple corollary of it involving Hadamard matrices.

Lemma 2.14** (Plotkin bound, special case).**

Let $V=\mathbb{F}_{2}^{k}$ and let $C$ be a finite subset of a subspace $W$ of dimension $m$. Let $d$ be the minimum distance in $C$, namely

[TABLE]

where $H(x,y)$ denotes the Hamming distance between $x$ and $y$, i.e., the number of coordinates equal to $1$ in $x+y$. Let $A_{2}(m,d)$ be the maximum size of a subset $S$ of some $m$-dimensional subspace of $V$ having minimum distance $d$. If $d$ is even and $2d>m$, then

[TABLE]

Lemma 2.15**.**

Let $H=(H_{i,j})$ be a Hadamard matrix of order $n>2$. If a submatrix $L$ of $H$ of size $a\times b$ contains only $1$’s (or only $-1$’s), then $ab\leq n$.

Proof.

Without loss of generality, assume that all entries of $L$ are equal to $1$. Let $1\leq i_{1},\ldots,i_{a}\leq n$ and $1\leq j_{1},\ldots,j_{b}\leq n$ be indices such that the entries $(L_{r,s})$ of $L$ with $1\leq r\leq a$ and $1\leq s\leq b$ satisfy

[TABLE]

Let $R_{i_{1}},\ldots,R_{i_{a}}$ be the corresponding rows of $H$ and $T_{j_{1}},\ldots,T_{j_{b}}$ the corresponding columns of $H$. Put $C=\{R_{i_{1}},\ldots,R_{i_{a}}\}$.

For any pair of rows $R_{i_{h}},R_{i_{k}}$ in $C$ (say $R_{i_{1}},R_{i_{2}}$), define: (a) $x_{1,2}$ as the number of columns $T_{k}$ with $k\notin\{j_{1},\ldots,j_{b}\}$ in which $H_{i_{1},k}=H_{i_{2},k}$; (b) $y_{1,2}$ as the number of such columns in which $H_{i_{1},k}\neq H_{i_{2},k}$. Thus,

[TABLE]

while the orthogonality of $R_{i_{1}}$ and $R_{i_{2}}$ implies

[TABLE]

From (15) and (16) we obtain

[TABLE]

In other words, (17) says that

[TABLE]

since all elements of $L$ are equal to $1$.

We have thus proved that the minimum Hamming distance $d$ in $C$ satisfies

[TABLE]

Let $m=n-b$. With the notation of Lemma 2.14, we have

[TABLE]

Since $H$ is Hadamard and $n>2$, the order $n$ is a multiple of $4$. Hence (19) implies that $d$ is even. Moreover, $2d=n>m=n-b$, so that (20) and Lemma 2.14 imply

[TABLE]

Multiplying (21) by $b$ we obtain

[TABLE]

This proves the result. ∎

3. Proof of Theorem 1.2

Proof.

By Lemma 2.1, it suffices to prove parts (c) and (d).

We first prove (c): By Lemma 2.5, we can assume that the first row of $E_{1}$ has exactly $h^{2}$ entries equal to $1$ and $h^{2}$ entries equal to $-1$. This implies that $E_{2}$ has $h^{2}+h$ entries in its first row equal to $1$ and $h^{2}-h$ entries equal to $-1$. On the other hand, either all entries in the first row of $E_{2}$ are equal, in which case we must have $h^{2}+h=2h^{2}$, i.e., $h^{2}=h$, hence $h=1$, or half of them are equal to $1$, in which case we must have either $h^{2}+h=h^{2}$ or $h^{2}-h=h^{2}$, both of which are impossible. Therefore, $h=1$, and hence $n=4$.

To prove part (d), put $R_{1}:=[k_{1},\ldots,k_{n/2}]$. Observe that $\langle R_{1},R_{1}\rangle=n/2$. Lemma 2.4, applied to each of the terms $k_{j}k_{j-1}$ that sum up to $\langle R_{1},R_{2}\rangle$ (for indices $j=1,\ldots,n/2+1$, with $k_{n/2+1}=k_{1}$), shows that all $k_{j}$ with odd $j$ are equal, say to $c_{1}\in\{-1,1\}$ (there are exactly $h^{2}$ such values of $j$), and that all $k_{j}$ with even $j$ are equal, say to $c_{2}\in\{-1,1\}$ (there are also exactly $h^{2}$ such values of $j$, since $n/2=2h^{2}$). By Lemma 2.5, we can assume without loss of generality that $c_{1}+c_{2}=0$. Thus, the first row of $E_{1}$ has exactly $h^{2}$ entries equal to $1$ and $h^{2}$ entries equal to $-1$.

Using the same argument with $S_{1}$, it remains to consider the following case: Lemma 2.5 implies that the first row of $E_{2}$ has $h^{2}$ entries equal to $1$ and $h^{2}$ entries equal to $1$, i.e., $c_{1}=c_{2}$. In other words, all entries of $E_{2}$ are equal to $1$. This implies that $E_{1}$ must contain $h$ entries equal to $1$ and $2h^{2}-h$ entries equal to $-1$. This is impossible when $h>1$ by Lemma 2.5 since $E_{1}$ has order $2h^{2}$ and rank $1$. Therefore, $h=1$, and hence $n=4$. ∎

4. Proof of Theorem 1.5

Proof.

Assume, by contradiction, that $H$ is a circulant Hadamard matrix with $n>4$.

We first prove part (a). By Lemma 2.10, we have $n=4h^{2}$ with $h$ odd. Assume, without loss of generality, that $E_{1}$ has rank $1$. By Lemma 2.5, we can assume, without loss of generality, either:

(i) $h_{1}=h_{3}=\ldots=h_{n-1}$, or (ii) $h_{1}=1$, $h_{3}=-1$, $\ldots$, $h_{n-3}=1$, $h_{n-1}=-1$.

When (i) holds, we consider the submatrix $L$ of $H$ equal to $E_{1}$. More precisely, the entries of $L$ are those of $E_{1}$, at the intersection of rows $1,3,5,\ldots,4h^{2}-1$ and columns $1,3,5,\ldots,4h^{2}-1$ of $H$. Since $L$ has size $2h^{2}\times 2h^{2}$ and all its entries are equal to $1$, Lemma 2.15 implies that

[TABLE]

In other words, (23) says that $h^{2}\leq 1$. But $h$ is odd, so $h=1$, and $n=4$, contradicting our assumption $n>4$. This proves part (i) of (a).

Similarly, when (ii) holds, we consider the submatrix $L$ of $H$ equal to the submatrix $F_{1}=\operatorname{circm}(h_{1},h_{5},\ldots,h_{4h^{2}-5},h_{4h^{2}-1})$ of $E_{1}$. The entries of $L$ are those of $F_{1}$ at the intersection of rows $1,5,\ldots,4h^{2}-5,4h^{2}-1$ and columns $1,5,\ldots,4h^{2}-5,4h^{2}-1$ of $H$. Since $L$ has size $h^{2}\times h^{2}$ and all its entries are equal to $1$, Lemma 2.15 implies that

[TABLE]

so that $h^{2}$ is an odd number with $h^{2}\leq 4$, i.e., $h=1$. As before, we get the contradiction $n=4$. This proves part (a) of the theorem.

To prove part (b), assume that both $E_{1}$ and $E_{2}$ have rank $2$. By Lemma 2.10, $n=4h^{2}$ with $h$ odd. Then, by Lemma 2.13 (and changing $H$ to $-H$ if necessary), we may assume that

[TABLE]

Let $v,w$ be the first two rows of $E_{2}=\operatorname{circm}(h_{2},h_{4},\ldots,h_{4h^{2}})$. We claim that $v$ and $w$ are $\mathbb{Q}$-linearly independent. Otherwise, putting $C=E_{2}$ in Lemma 2.5 yields

[TABLE]

From (26) and (25) we get

[TABLE]

i.e., the contradiction $h=1$. This proves the claim.

Let $u$ be the third row of $E_{2}$. Since $v$ and $w$ form a basis of the $\mathbb{Q}$-vector space generated by the rows of $E_{2}$, there exist rational numbers $a,b$ such that

[TABLE]

Let $S=h_{2}+h_{4}+\cdots+h_{4h^{2}}$. Since, by (25), $S\neq 0$, Lemma 2.6 with $D=E_{2}$ and $s=S$ implies

[TABLE]

Let $h_{2j}$ be any entry of $u$, with $j=2h^{2}-1,2h^{2},1,2,\ldots,2h^{2}-2$. By (28) we have

[TABLE]

Assume $ab\neq 0$. By Lemma 2.7, (30) implies

[TABLE]

for all such $j$. In other words, all entries of $v$ are equal. By (25) they must be $1$, and so (25) reads exactly as (27), giving again the contradiction $h=1$. Therefore $ab=0$. By (29), the only cases in (28) are $w=u$ when $(a,b)=(0,1)$ and $v=u$ when $(a,b)=(1,0)$.

Since $w$ and $u$ are consecutive rows of $E_{2}$, the same argument used above to show $v$ and $w$ are $\mathbb{Q}$-linearly independent applies to show $w$ and $u$ are also $\mathbb{Q}$-linearly independent. Thus, $w\neq u$, so the case $w=u$ is impossible. Assume then $v=u$. By Lemma 2.9 with $C=E_{2}$ and $s=S$, we obtain

[TABLE]

But by (25), $S=2h$. From (32) it follows that (27) still holds, yielding again the contradiction $h=1$.

This finishes the proof of the theorem. ∎