Aspects of Optimality of Plans Orthogonal Through Other Factors and Related Multiway Designs

Abstract

In a blocked main effect plan (MEP), a pair of factors is said to be orthogonal through the block factor if their totals adjusted for the block are uncorrelated, as defined in Bagchi (Technometrics 52:243–249, 2010). This concept is extended here to orthogonality through a set of other factors. We discuss the impact of such an orthogonality on the precision of the estimates as well as on the data analysis. We construct a series of plans in which every pair of factors is orthogonal through a given pair of factors. Next we construct plans orthogonal through the block factors (POTB). We construct the following POTBs for symmetrical experiments. There are an infinite series of E-optimal POTBs with two-level factors and an infinite series of universally optimal plans for three-level factors. We also construct an universally optimal POTB for an $s^t(s+1)$ experiment on blocks of size $(s+1)/2$, where $s \equiv 3 \pmod 4$ is a prime power. Next we study optimality aspects of the “duals” of main effect plans with desirable properties. Here by the dual of a main effect plan we mean a design in a multi-way heterogeneity setting obtained from the plan by interchanging the roles of the block factors and the treatment factors. Specifically, we take up two series of universally optimal POTBs for symmetrical experiments constructed in Morgan and Uddin (Ann Stat 24:1185–1208, 1996). We show that the duals of these plans, as multi-way designs, satisfy M-optimality. Finally, we construct another series of multiway designs, which are also duals of main effect plans, and proved their M-optimality. This result generalizes the result of Bagchi and Shah (J Stat Plan Inf 23:397–402, 1989) for a row–column set-up. It may be noted that M-optimality includes all commonly used optimality criteria like A-, D- and E-optimality.

Appendices

Appendix A : Results involving linear algebra

The proofs of the results in Section 3 :

The results in the following Lemma are well-known. [See Exercise 7 of Chapter 2 of Yanai, H., Takeuchi, K. and Takane, K. [35], for instance].

Lemma 8.1

Consider matrices U, V, W with the same number of rows. Then, the following hold.

• (a) Suppose $\mathcal{C} (V) \subseteq \mathcal{C} (W)$. Then

  $$\begin{aligned} \mathcal{C} (P_V U) = \mathcal{C} (P_W U) \Leftrightarrow (P_W - P_V)U =0. \end{aligned}$$

• (b) If $W = [U,V]$ then $P_W - P_V = P_Z$, where $Z = (I - P_V) U$.

Proof of Theorem 3.1:

Taking $W = X_{{\bar{i}}}, \; U = X_S$ and $V = X_T$ and applying Lemma 8.1 we find that

$$\begin{aligned} P_{{\bar{i}}} = P_T + P_Z, \text{ where } Z = (I - P_T) X_S. \end{aligned}$$

(8.1)

Proof of (a): In view of (2.3), (2.5) and (8.1) we see that $C_{i;{\bar{i}}} = C_{i;T} - X'_i P_Z X_i$. Therefore, the required necessary and sufficient condition is that $P_Z X_i = 0$, which is equivalent to (3.1). Hence the result.

Proof of (b): In view of Lemma 2.2 the following hold.

$$\begin{aligned} SS_{i;{\bar{i}}}= & {} Y^{\prime }P_U Y, SS_{i;T} = Y^{\prime }P_V Y, \end{aligned}$$

(8.2)

$$\begin{aligned} \text{ where } U= & {} (I - P_{{\bar{i}}}) X_i, V = (I - P_T)X_i. \end{aligned}$$

(8.3)

Since the support of Y is $R^n$, $Y^{\prime }P_U Y = Y^{\prime }P_V Y$ w.p.1 if and only if $P_U = P_V$. Thus, the required necessary and sufficient condition is that $\mathcal{C} (U) = \mathcal{C} (V)$. Applying Lemma 8.1 we see that the required necessary and sufficient condition is

$$\begin{aligned} ( P_{{\bar{i}}} - P_T) X_i = 0. \end{aligned}$$

(8.4)

But in view of (8.1) this is the same as $P_Z X_i = 0$, which is equivalent to (3.1). Hence the result. $\square$

The proofs of the results in Section 5 :

Proof of Lemma 5.2: Eliminating ${\hat{\mu }}$ from the normal equations [see Lemma 2.1] we get a system of equation in $\hat{\beta _i}, \; i \in \mathcal{M}$ and ${\hat{\tau }}$. These equations are as follows.

$$\begin{aligned}&\sum \limits _{ i,j \in \mathcal{M}} C_{ij;0} \hat{\beta _j} + C_{iV;0} {\hat{\tau }} = {{\mathbf {Q}}}_{i;0}, \; i \in \mathcal{M} \end{aligned}$$

(8.5)

$$\begin{aligned}&\text{ and } \sum \limits _{i \in \mathcal{M}} C_{Vi;0} \hat{\beta _i} + C_r {\hat{\tau }} = {{\mathbf {Q}}}_{V;0}. \end{aligned}$$

(8.6)

Here $C_{ij;0}$’s and ${{\mathbf {Q}}}_{i;0}$’s are as in Notation 2.3 (c).

Proof of (a): By the hypothesis, we find that the individual C-matrices involving only the blocking factors are as follows.

$$\begin{aligned} C_{ii';0} = \left\{ \begin{array}{ll} (p+s)K_s &{} \text{ if } i' = i, \\ pK_s &{} \text{ otherwise } \end{array} \right. ,\; i,i'\in \mathcal{M}, \end{aligned}$$

(8.7)

Here $p = -1$ for the type 1 setting and $K_s$ is as in (6.6)

For a fixed i, we eliminate all $\hat{\beta _i'} , i' \ne i$ from (8.5) by using (8.7). Then we get an equation involving only $\hat{\beta _i}$ and ${\hat{\tau }}$. We use this equation to eliminate all $\hat{\beta _i}$’s from (7.24). Then we get the reduced normal equation for ${\hat{\tau }}$, in the form $C_d {\hat{\tau }} = Q$, where $C_d$ is as in the statement.

Proof of (b):

By the hypothesis, the individual C-matrices are as follows.

$$\begin{aligned} C_{ii';0}= & {} \left\{ \begin{array}{ll} (s+1)K_s &{} \text{ if } i' = i, \\ K_s &{} \text{ otherwise } \end{array} \right. ,\; 1 \le i,i' \le m, \end{aligned}$$

(8.8)

$$\begin{aligned} C_{i\infty ;0}= & {} \left\{ \begin{array}{ll} sK_{s+1} &{} \text{ if } i = \infty , \\ 0 &{} \text{ otherwise } \end{array} \right. ,\; 1 \le i,i' \le m. \end{aligned}$$

(8.9)

Following the same procedure as in Case (a) we get the C-matrix as in the statement. $\square$

The proofs of the results in Section 6 :

Proof of Lemma 6.7: Let $X = \{x_1, \cdots x_{s-1}\}$ be a set of orthonormal vectors in $\langle 1_s \rangle ^{\perp }$. Let $Z = \{ z_i = 1_h \otimes x_i, \; x_i \in X\}$. For a $z \in Z, \; Nz = Hx_i$ for some i. So, by Lemma 6.6 (b), $Nz = -z$, so that $z'N'Nz = z'z$. Let ${\tilde{\mu }}_1 \le \cdots \le {\tilde{\mu }}_{s-1}$ be the smallest $s-1$ positive eigen values of $NN'$. Then, ${\tilde{\mu }}_1, \cdots , {\tilde{\mu }}_{s-1}$ are also the smallest $s-1$ positive eigen values of $N'N$. Therefore,

$$\begin{aligned} \sum _{i=1}^{s-1} {\tilde{\mu }}_i \le \sum _{i=1}^{s-1} (z'_i N'N z_i)/(z'_i z_i) = (s-1). \end{aligned}$$

(8.10)

Since N is an incidence matrix, the largest eigenvalue of $NN'$ corresponds to the eigenvector $1_v$. Thus, the eigenvector $e_i$ corresponding to ${\tilde{\mu }}_i$ cannot be $1_v$ and therefore $e'_i 1_v = 0$, for each i. Hence the result follows from (6.5) and (8.10) . $\Box$

Proof of Lemma 6.8: Let $\mathcal{N} (A)$ denote the null space of the matrix A and $\nu (A)$ the nullity of A.

By (6.9)

$$\begin{aligned} C_d = rK_v - \frac{1}{s} E_\mathcal{B} E'_\mathcal{B} - \frac{1}{s(s-h)} S_{\mathcal{B}} S'_{\mathcal{B}}. \end{aligned}$$

By Lemma 5.1 (a) and the definition of $S_\mathcal{B}$, $\;1'v E_\mathcal{B} = 0 = 1'_v S_\mathcal{B}$. Thus, $1'v C_d = 0$ and (0) is proved.

We now prove (i). Substituting for $E_\mathcal{B} E'_\mathcal{B}$ from (6.5) and $S_\mathcal{B} S'_\mathcal{B}$ from (6.6) we get the following expression for $C_d$.

$$\begin{aligned} C_d = r K_v - \frac{1}{s}(NN' - \frac{r(s-1)}{s}J_v) - \frac{1}{s(s-h)} (HH' - \frac{(s-1)^2}{s}J_v). \end{aligned}$$

By definition of $H = H_d$, $\mathcal{N} (NN') \subset \mathcal{N} (HH')$. Now, let $W = \langle 1_h \rangle ^{\perp } \otimes 1_s$. Since d is equireplicate, $Nw = 0, \forall w \in W$. Therefore, $\nu (NN') = \nu (N'N) \ge h-1$, implying $\nu (HH') \ge h-1$. Let $x \in \mathcal{N} (NN')$. Since $1_v$ is the eigenvector of $NN'$ corresponding to the largest eigenvalue, $x \ne 1_v$. Therefore, $x' 1_v = 0$. So, $C_d x = r$. Hence (i) follows.

Next we proceed to prove (ii), which is about the next largest eigen values of $C_d$, Let $P = a E_\mathcal{B} E'_\mathcal{B} + b S_\mathcal{B} S'_\mathcal{B} , \; a,b >0$. While proving (i), we have also proved that $\mu _i (P) = 0,\; 0 \le i \le h-1$. Let ${\tilde{\mu }}_1 (T) \le \cdots \le {\tilde{\mu }}_{s-1} (T)$ be the smallest $s-1$ positive eigen values of $T,\; T = P$ or $E_\mathcal{B} E'_\mathcal{B}$. Fix $i : 1 \le i \le s-1$. By an well-known result [see Corollary III.2.2 of Bhatia [9], for instance] we get

$$\begin{aligned} {\tilde{\mu }}_i (P) \le a {\tilde{\mu }}_i (E_\mathcal{B} E'_\mathcal{B}) + b \mu _{v-1} (S_{\mathcal{B}} S'_{\mathcal{B}}), \end{aligned}$$

which is $a {\tilde{\mu }}_i (E_\mathcal{B} E'_\mathcal{B}) + bh$, by Lemma 6.6 (c). So, by Lemma 6.7

$$\begin{aligned} \sum _{i= 1}^{s-1} {\tilde{\mu }}_i (P) \le (s-1) (a + bh). \end{aligned}$$

(8.11)

But P becomes $r K_v - C_d$, if we put $a = \frac{1}{s}, b = \frac{1}{s(s-h)}$. So, the result follows from (8.11). $\square$

The proof of Theorem 6.2 :

We begin with an well-known result.

Lemma 8.2

Suppose $x_1, \cdots x_n$ are real numbers with $\sum _{i=1}^{n} x_{i} = a$. Then, the following hold. (a) $\sum _{i=1}^{n} x^2_{i} \ge a^2/n, = \text{ when } x_i = a/n, \; \forall i$.

(b) In particular, if $x_i$’s are integers, then $\sum _{i=1}^{n} x^2_{i}$ is minimum, when $x_i = [a/n]$ or $[a/n] + 1$.

Proof of Lemma 6.9:

• (a) Follows by straightforward computation.

• (b) Fix an arbitrary $x \in R^p$. For $1 \le i \le k$, let us write $A'_ix$ as $y_i = [\begin{array}{cccc} y_{i1}&\cdots y_{iq} \end{array}]^\prime$ . Then, $x' \sum _{i=1}^k A_i A'_i x = \sum _{i=1}^k y'_i y_i = \sum _{j=1}^q \sum _{i=1}^k y^2_{ij}$. Now, $\sum _{i=1}^k y_{ij} = \sum _{l=1}^p T(l,j) x_l = z_j$, say. Therefore, by Lemma 8.2, $\sum _{i=1}^k y^2_{ij} \ge z^2_j /k$. Since $z_j$ is the jth entry of $T'x$, the result follows.

• (c) $Tr(HH') = \sum _{i=1}^{p} \sum _{j=1}^{q} (H(i,j))^2.$ Again, $\sum _{i=1}^{p} \sum _{j=1}^{q} H(i,j) = kpr$. Therefore, by Lemma 8.2, $\sum _{i=1}^{p} \sum _{j=1}^{q} (H(i,j))^2$ is minimum, when the following hold. $H (i,j) = [rk/q] \text{ or } [rk/q] + 1, 1 \le i \le p, 1 \le j \le q.$ Since a sufficient condition for the above is $A_1 = A_2 = \cdots A_k, \text{ and } A_1 (i,j) = 0 \text{ or } 1, \;\; 1 \le i \le p, 1 \le j \le q$, the result follows. $\square$

Proof of Lemma 6.10:

By definition of $C, \; \mu _0(C) = 0 = \gamma _0, \; \; \mu _i (C) = d - \mu _{n+1-i} (A)$ for $i > 0$. Therefore, $\sum _{i=1}^{\rho } \mu _{i} (C) = d \rho - tr(A) \le \rho (d - a)$. Since $(1/l)\sum _{i=1}^l \mu _{i} (C)$ is increasing in l, it follows that $\sum _{i=1}^l \mu _{i} (C) \le l(d - a), \; 1 \le l \le \rho$. Since $\mu _i(C) \le d \; \forall i$, the result follows. $\square$

Proof of Lemma 6.12:

Since d is equireplicate, $1_v$ is an eigenvector of $T_i$ as well as of $T_j$ with eigenvalue r, where r is the replication number (for the treatments). Again, by the hypothesis, $T_i T_j = r^3 J_v = T_j T_i$. Thus, $T_i$ and $T_j$ are commuting matrices and hence there is an orthonormal basis consisting of common eigenvectors of these two matrices. Therefore, $\mathcal{C} (T_i) \cap \mathcal{C} (T_i) = \mathcal{C} (T_i T_j) = \mathcal{C} (J_v) = \langle \{1_v\} \rangle$. Hence the result. $\square$

Appendix B: Results involving finite field

The proofs of the results in Section 4 : We begin with the proof of Theorem 4.2.

Let $C_0$ be the set of all nonzero squares of F and $C_1$ the set of all nonzero non-squares of F.

Notation 9.1

Consider an $m \times n$ array P with entries from F. For $1 \le i,j\le m, \; k \in F$, let

$$\begin{aligned} d^k_{ij} = |\{l : p_{jl} - p_{il} = k \}|. \end{aligned}$$

$C_0 P$ will denote the $m \times (s-1)n/2$ array $\{cP: c \in C_0\}$.

Lemma 9.1

Suppose $\exists$ an array P as in Notation 9.1satisfying the following. $\sum \limits _{ k \in C_0} d^k_{ij} = \sum \limits _{ k \in C_1} d^k_{ij} = u_{ij}$, (say). Let $\mathcal{P}$ be a plan having the set of columns of the array $C_0 P$ as the set of runs. Let $\mathcal{P}^* = \mathcal{P} \oplus F$. Then, for a pair $(i,j), i\ne j,\;1 \le i,j \le m$, the incidence matrix $N_{ij}$ of $\mathcal{P}^*$ satisfies the following.

$$\begin{aligned} N_{ij} (x,y) = w _{ij}I_s + u_{ij} J_s, \text{ where } w _{ij} = (s-1)n/2 - su_{ij}. \end{aligned}$$

Proof

$N_{ij} (x,y) = |\{(l,\alpha ,q) : q \in C_0,\; q(p_{jl} - p_{il}) = y-x, \alpha = x - qp_{il}, \; 1 \le l \le n \}|$, which is $=|\{(l,q) : y-x = q(p_{jl} - p_{il}), \; q \in C_0, 1 \le l \le n \}|$. So, by hypothesis,

$$\begin{aligned} \text{ if } y-x \in C_0\text{, } \text{ then } N_{ij} (x,y) = |\{ l : p_{jl} - p_{il} \in C_0\}| = \sum \limits _{ k \in C_0} d^k_{ij} = u_{ij}. \end{aligned}$$

Similarly, if $y-x \in C_1$, then $N_{ij} (x,y) = \sum \limits _{ k \in C_1} d^k_{ij}$. Therefore, if $y = x$, then $N_{ij} (x,y) = \frac{s-1}{2}(n - 2 u_{ij})$. Hence the result. $\square$

Using the fact that when $s \equiv 3 \pmod 4$, $-1 \in C_1$, we can prove the following result from Lemma 9.1.

Lemma 9.2

Suppose $s \equiv 3 \pmod 4$ is a prime power and n is a multiple of 4. Suppose there is an $m \times n$ array P with entries $0,1,-1$ (viewed as members of F) satisfying the following.

(a) $p_{1,j} = 0, 1 \le j \le n$.

(b) For every ordered pair (i, j), $p_{il} - p_{jl} \in \{0,1,-1\}, \; 1 \le l \le n$.

(c) For $k = 1,-1, d^k_{ij} = \left\{ \begin{array}{ll} n/2 &{} \text{ if } (i,j) = (1,2), \\ n/4 &{} \text{ otherwise } \end{array} \right.$

Then the plan $\mathcal{P}^* = C_0 \mathcal{P} \oplus F$ satisfies the conditions of Theorem 4.1 with $c = ns(s-1)/8$. Here $C_0 \mathcal{P} \oplus F$ is to be interpreted as in Notation 4.1 (d)

Proof of Theorem 4.2 : Let H be a Hadamard matrix of order q. W.l.g., we assume that the first row of H consists of only 1’s. Write $H = [\begin{array}{cc} 1_q&{\tilde{H}} \end{array}]'$. Consider the array

$$\begin{aligned} P = \left[ \begin{array}{ccc} 0_{1 \times q} &{}|&{} 0_{1 \times q} \\ J_{1 \times q} &{}|&{} - J_{1 \times q} \\ ({\tilde{H}} + J_{q-1 \times q} )/2 &{}|&{} - ({\tilde{H}} + J_{q-1 \times q} )/2\\ ({\tilde{H}} + J_{q-1 \times q} )/2 &{}|&{} ({\tilde{H}} - J_{q-1 \times q} )/2\\ \end{array} \right] . \end{aligned}$$

It is easy to check that P satisfies the conditions of Lemma 9.2. Hence the proof of Theorem 4.2 is complete in view of Theorem 4.1. $\square$

The proofs of the results in Section 6.1 :

We present the result in Lemma 6.2 (c) as a theorem.

Theorem 9.1

The spectrum of $C_{d^*_1}$ is $r^{h-1} (r - (1/(s-h))^{s-1} (r-1)^{(h-1)(s-1)}.$

This is the only part of this paper where matrices with complex entries occur. If A is such a matrix, then $A^*$ will denote its conjugate transpose. To prove this theorem, we need a number of tools. We introduce some notations. Some of these were already in the main body of the paper; we have repeated them for the sake of readability.

Notation 9.2

(0) $s = p^m = ht+1$, where p is a prime, m, t, h are integers, $m \ge 1, h,t \ge 2$. $F_p$ and $F_s$ are finite fields of orders p and s , respectively. [In this section, we use the notation $F_s$ (rather than F like in the other sections) so as to distinguish it from the field of order p]

(i) Addition and subtraction in $I_h = \{0.1, \cdots h-1\}$ will always be modulo h. $C_i, \; i \in I_h$ will denote the cosets of the subgroup $C_0$ of order t in $F_s^*$, ordered in such a way that $C_iC_j=C_{i+j}$ for $i,j \in I_h$.

The rows and columns of every $s \times s$ (respectively, $h \times h$) matrix will be indexed by $F_s$ (respectively, $I_h$). Moreover, the rows and columns of every $hs \times hs$ matrix will be indexed by $I_h \times F_s$.

(ii) As in Notation 4.6, M will denote the $hs \times hs$ matrix $(( M_{i-j}))_{i,j \in I_h}$, where for $i \in I_h, M_i$ will denote the $s \times s$ matrix given by $M_i(x,y) = \left\{ \begin{array}{ll} 1 &{} \text{ if } y - x \in C_i, \\ 0 &{} \text{ otherwise } \end{array} \right.$

(iii) $\eta$ and $\omega$ are primitive h th and p th roots of unity, respectively.

(iv) Consider the function trace $: F_s \rightarrow F_p$ defined as follows. $trace(x) = \sum _{i=1}^{m} x^{p^i},\; x \in F_s$. [This is $F_p$-linear and into $F_p$ since $x \rightarrow x^p$ is an automorphism of $F_s$ and its fixed field is $F_p$].

(v) U and V are unitary matrices of orders h and s , respectively, given as follows.

$$\begin{aligned} U(i,j) = (1/\sqrt{h}) \eta ^{ij}, i,j \in I_h \text{ and } V(x,y) = (1/\sqrt{s}) \omega ^{trace(xy)}, x,y \in F_s. \end{aligned}$$

(vi) Consider the sums $g_i$ given by $g_i = \sum \limits _{ x \in C_i} \omega ^{-trace(x)}, i \in I_h$.

(vii) For $k \in I_h$, $G_k$ is the $h \times h$ matrix given by $G_k(i,j) = g_{i-j+k}, \; i,j \in I_h$.

(viii) For $k \in I_h$, $E_k$ is the $s \times s$ diagonal matrix given by

$$\begin{aligned} E_k(x,x) = \left\{ \begin{array}{ll} - t &{} \text{ if } x =0 \\ 1 &{} \text{ if } x \in C_k \\ 0 &{} \text{ otherwise } \end{array} \right. \end{aligned}$$

(ix) W will denote the $hs \times hs$ unitary matrix $U \otimes V$.

(x) T is the $h \times h$ diagonal matrix with the entries: $T(l,l) = \eta ^l, \; l \in I_h$0

We study the behaviour of M under the actions of U and V.

Lemma 9.3

$(I_h \otimes V)^* M (I_h \otimes V) = \sum \limits _{ k \in I_h} G_k \otimes E_k$.

Proof

A computation shows that we have $V^* M_i V = Diag(\lambda _i(x),\; x \in F_s) = \sum \limits _{ k \in I_h} g_{i+k} E_k,$ where

$$\begin{aligned} \lambda _i (x) = \left\{ \begin{array}{ll} t &{} \text{ if } x =0 \\ g_{i+k} &{} \text{ if } x \in C_k, k \in I_h.\end{array} \right. \end{aligned}$$

[To verify the second equality here (as well as for later use), we need to observe that

$$\begin{aligned} \sum \limits _{ k \in I_h} g_{i+k} = \sum \limits _{ k \in I_h} g_k = \sum \limits _{ x \in F^*_s} \omega ^{-trace(x)} =-1, \text{ since } \sum \limits _{ x \in F_s} \omega ^{-trace(x)} = 0]. \end{aligned}$$

$\square$

But, by the definition of M, the left hand side of the statement of this lemma is $((V^* M_{i-j} V))_{i,j \in I_h}$. Hence the result follows from the definition of $G_k$ and $E_k$.

Lemma 9.4

$W^* M W$ is a diagonal matrix with the following entries. For $i \in I_h, x \in F_s$, the (i, x)th diagonal entry of $W^* M W$ is

$$\begin{aligned} \delta (i,x) = \left\{ \begin{array}{ll} s-1 &{} \text{ if } \;\; i=0, x =0, \\ 0 &{} \text{ if } \;\; i \ne 0, x =0, \\ \eta ^{ik}\sum \limits _{ j \in I_h} g_j \eta ^{-ij} &{} \text{ if } \;\; x \in C_k. \end{array} \right. \end{aligned}$$

Proof

Since $W = (I \otimes V) (U \otimes I)$, Lemma 9.3 implies that

$$\begin{aligned} W^* M W = (U \otimes I)^* \left( \sum \limits _{ k \in I_h} G_k \otimes E_k \right) (U \otimes I) = \sum \limits _{ k \in I_h} (U^* G_k U) \otimes E_k. \end{aligned}$$

But one can verify that

$$\begin{aligned} U^* G_k U = \sum \limits _{ j \in I_h} g_{k-j} T^j, \; k \in I_h. \end{aligned}$$

So, $W^* M W = \sum \limits _{ k \in I_h} \sum \limits _{ j \in I_h} g_{k-j} T^j \otimes E_k$. Now, the formulae for T and $E_k$ imply the result. $\square$

Notation 9.3

(a) $\Omega _h$ is the multiplicative group of all hth roots of unity.

(b) For $i \in I_h$, $\chi _i : F^*_s \rightarrow \Omega _h$ is defined by $\chi _i (x) = \eta ^{-ij}$, if $x \in C_j, j \in I_h$.

Lemma 9.5

$|\sum \limits _{ j \in I_h} g_i \eta ^{-ij}|^2 = \left\{ \begin{array}{ll} 1 &{} \text{ if } i=0,\\ s &{} \text{ if } 0< i < h \end{array} \right.$

Proof

Since $C_jC_k = C_{j+k}$ (where the addition in the suffix is modulo h), $\chi _i$ is a group homomorphism (character) on $F^*_s$. From the definition of $g_j$’s we have

$$\begin{aligned} \sum \limits _{ j \in I_h} g_j \eta ^{-ij} = \sum \limits _{j \in I_h} \sum \limits _{x \in C_j} \omega ^{-trace(x)} \chi _i(x) = \sum \limits _{x \in F_s^*} \omega ^{-trace(x)}\chi _i(x) = g(\chi _i), \end{aligned}$$

which is the Gauss sum attached to the character $\chi _i$. But $|g(\chi _i)|^2 = \left\{ \begin{array}{ll} 1 &{} \text{ if } i=0,\\ s &{} \text{ if } i \ne 0 \end{array} \right.$, by a classical result on such Gauss sums [see, for instance Chapter 10 of Ireland and Rosen [22]]. Hence the result. $\square$

Putting the information from Lemmas 9.4 and 9.5 together, we get the spectrum of $M M'$.

Lemma 9.6

$W^* M M' W= D$, where the diagonal entries of the diagonal matrix D are as follows.

$$\begin{aligned} |\delta (i,x)|^2 = \left\{ \begin{array}{ll} (s-1)^2 &{} \text{ if } i=0, x =0 \\ 0 &{} \text{ if } i \ne 0, x =0 \\ 1 &{} \text{ if } i=0, x \ne 0, \\ s &{} \text{ if } i \ne 0, x \ne 0. \end{array} \right. \end{aligned}$$

In order to get the spectrum of $C_{d_1^*}$ we need the spectrum of $HH'$.

Lemma 9.7

$W^* HH' W$ is a diagonal matrix with the (i, xth entry $= \left\{ \begin{array}{ll} h(s-1)^2 &{} \text{ if } i=0, x =0 \\ h &{} \text{ if } i \ne 0, x \ne 0 \\ 0 &{} \text{ if } i \ne 0. \end{array} \right.$

Proof

Let $\triangle _h$ denote the $h \times h$ matrix having the (0, 0)th entry 1 and all other entries 0. $\triangle _s$ is defined in a similar manner. It is easy to verify that

$$\begin{aligned} U^* J_h U = h \triangle _h \text{ and } V^* J_s V = s \triangle _s. \end{aligned}$$

Since $H = 1_h \otimes ( J_s - I_s )$ [recall Lemma 6.6], we see that

$$\begin{aligned} W^* HH' W = h \triangle _h \otimes (s \triangle _s - I_s)^2, \end{aligned}$$

which is a diagonal matrix with the entries as in the statement. $\square$

Proof of Theorem 9.1 : Lemmas 9.6 and 9.7 imply the result in view of the expression for $C_{d^*_1}$ in Lemma 6.2. $\square$

The proofs of the results in Section 6.3 :

The following result lies at the foundation of the construction of $d^*_3$ :

Lemma 9.8

Let $s \equiv 3 \pmod 4$ be a prime power. Then there is a subset W of $C_0$ and a function $f : W \rightarrow C_1$ satisfying the following.

(a) $|W| = (s-3)/4$.

(b) For every $\xi \in W$, $(\xi - 1) (f(\xi ) - 1) \in C_0$.

(c) For $\xi \ne \xi ' \in W$, $(\xi - \xi ') ( f(\xi )- f(\xi ')) \in C_0.$

Proof

Let $W = \{x \in C_0 : 1-x^2 \in C_0\}, \; {\tilde{W}} = \{x \in C_1 : 1-x^2 \in C_1\}$. Note that $x \rightarrow -x$ is a bijection from W onto $(C_1 \setminus {\tilde{W}})\setminus \{-1\}$. Therefore, $|W| = |C_1 \setminus {\tilde{W}}| -1 = (s-3)/2 - |{\tilde{W}}|$. Thus, $|W| + |{\tilde{W}}| = (s-3)/2$. Also, $x \rightarrow -1/x$ is a bijection from W onto ${\tilde{W}}$. Thus, $|W| = |{\tilde{W}}|$, which proves (a).

Let $f : W \rightarrow {\tilde{W}}$ be defined by $f(x) = -1/x,\; x \in W$. Then, for $\xi \in W$, $1-\xi ^2 \in C_0$, so that $(1 -\xi ) (1 - f(\xi )) = \xi ^{-1} (1 - \xi ^2)$ which is in $C_0$. This proves (b).

Again, for $\xi \ne \xi ' \in W$, $(f(\xi ) - f(\xi '))(\xi - \xi ') = (\xi \xi ')^{-1} (\xi - \xi ')^2 \in C_0$, which implies (c). $\square$

Proof of Lemmma 6.11 : Recall (6.10). For $l = 0,1$, Let $\tilde{R^3_l}$ denote the set of $(w+2) \times 1$ vectors obtained from $R^3_l$ by juxtaposing 0 in the $(w+2)$th position of each member of $R^3_l$. Let $\mathcal{P}^0_l$ denote the plan for an $s^{w+1} (s+1)$ experiment with $\tilde{R^3_l}$ as the set of runs and the set of factors $\{A_1, \cdots A_w\} \cup \{A_\infty \} \cup \{A_0\}$. Let $\mathcal{P}_l = \mathcal{P}^0_l \oplus F$ and $\mathcal{P} = \bigcup \limits _{l =0,1} \mathcal{P}_l$. Needless to mention that while generating $\mathcal{P}_l$ from $\mathcal{P}^0_l$, we make use of (4.4).

Proof of (a): Fix $i \ne j, \; 1 \le i,j \le w$. Let $D_l$ denote the ($A_i$, $A_j$) plot difference for the plan $\mathcal{P}^0_l, l = 0,1$ [recall Definition 4.5]. From (6.10) we see that

$$\begin{aligned} D_0 = (\xi _j - \xi _i) {\bar{C}}_0 \text{ and } D_1 = (f(\xi _j) - f(\xi _i)) {\bar{C}}_1. \end{aligned}$$

By (c) of Lemma 9.8, $D_0 \sqcup D_1 = F \sqcup \{0\}$. Now applying Lemma 4.2 on the plan $\mathcal{P}^0_0 \cup \mathcal{P}^0_1$ we see that

$$\begin{aligned} L_{ij} = I_s + J_s. \end{aligned}$$

(9.1)

Next fix $i, 1 \le i \le w$. Let $D_l$ denote the ($A_i$, $A_\infty$) plot difference for the plan $\mathcal{P}^0_l, l = 0,1$. From (6.10) and the equation next to it we see that

$$\begin{aligned} D_0 = (1 - \xi _i) {\bar{C}}_0 \text{ and } D_1 = (1 - f(\xi _i))C_1 \cup \{\infty \}, \text{ so } \text{ that } D_0 \sqcup D_1 = F^+, \end{aligned}$$

by (b) of Lemma 9.8. Thus, by Lemma 4.2$L_{i\infty } = J_{s\times s+1}$. This relation together with (9.1) completes the proof.

Proof of (b): Fix $i, 1 \le i \le w$. Let $D_l$ be the ($A_0$,$A_i$) plot difference for the plan $\mathcal{P}^0_l, l = 0,1$. From (6.10) we see that

$$\begin{aligned} D_0 = \xi _i {\bar{C}}_0 \text{ and } D_1 = f(\xi _i) {\bar{C}}_1, \text{ so } \text{ that } D_0 = D_1 = {\bar{C}}_0, \end{aligned}$$

by the definition of W. Thus, $N^{(l)}_{0i} = M_0 + I_s,\; l = 0,1$, by Lemma 4.2. But from the description of $d^*_3$, in view of Notation 6.1 we see that

$$\begin{aligned} N_i = \left[ \begin{array}{l} N_{i0} \\ N_{i1} \end{array} \right] , \text{ where } N_{il} = N^{(l)}_{0i},\; i \in \{1, \cdots w\} \cup \{\infty \}. \end{aligned}$$

(9.2)

Hence the result.

Proof of (c): Now we consider the ordered pair of factors $(A_0,A_\infty )$ of $\mathcal{P}^0_l, l = 0,1$ and find the ($A_0$, $A_\infty$) for each of $\mathcal{P}_0$ and $\mathcal{P}_1$. We see that $D_0 = {\bar{C}}_0 \text{ and } D_1 = C_1 \cup \{\infty \} .$ Thus,

$$\begin{aligned} N^{(0)}_{0, \infty } = [ \begin{array}{ll} M_0 + I_s&0_{s \times 1} \end{array}], \text{ while } N^{(1)}_{0, \infty } = [ \begin{array}{ll} M_1&1_s \end{array}]. \end{aligned}$$

Now using (9.2) we get the result. $\square$