几类n2阶BCCB哈达码矩阵的构造On the Construction of Several Types of BCCB Complex Hadamard Matrices of Order n2

• 全文下载: PDF(477KB)    PP.656-661   DOI: 10.12677/PM.2018.86088
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In this note, we study how to construct BCCB complex Hadamard matrices. We first give a necessary and sufficient condition for a BCCB complex matrix of order n2 to be Hadamard, and then use the condition to construct various types of BCCB complex Hadamard matrices. As an example, three new types of BCCB complex Hadamard matrices of order 16 are provided.

1. 引言

2. 预备知识

$n\ge 2$ 是自然数。

1) 对角阵 $D=diag\left({a}_{0},{a}_{1},\cdots ,{a}_{n-1}\right)$ 定义如下：

$\forall 1\le i,j\le n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{jk}={\delta }_{jk}{a}_{k-1}$ .

2) 循环阵 $C=Circ\left({a}_{0},{a}_{1},\cdots ,{a}_{n-1}\right)$ 定义如下：

$\forall 1\le j,k\le n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{j,k}={a}_{\left(n-1\right)j\oplus k}$

$C=\left(\begin{array}{ccccc}{a}_{0}& {a}_{1}& {a}_{2}& \cdots & {a}_{n-1}\\ {a}_{n-1}& {a}_{0}& {a}_{1}& \cdots & {a}_{n-2}\\ {a}_{n-2}& {a}_{n-1}& {a}_{0}& \cdots & {a}_{n-3}\\ \cdots & \cdots & \cdots & \ddots & \cdots \\ {a}_{1}& {a}_{2}& {a}_{3}& \cdots & {a}_{0}\end{array}\right)$

3) n2阶方阵C被称为由循环块组成的循环矩阵(BCCB矩阵)是指它具有下面形式：

$C=Circ\left({C}_{0},{C}_{1},\cdots ,{C}_{n-1}\right)$

4) 一个n阶方阵H被称为复哈达码矩阵是指：方阵中每个元素的模都是1，并且 ${H}^{*}H={I}_{n}$ ，其中， ${I}_{n}$ 是n阶单位阵，*是厄米特转置。

${\omega }_{n}={\text{e}}^{\frac{2\pi i}{n}}$ ，傅立叶矩阵 ${F}_{n}$ 定义如下：

$\forall 1\le i,j\le n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left({F}_{n}\right)}_{jk}={\omega }^{\left(j-1\right)\left(i-1\right)}$

${D}_{n}=diag\left(1,{\omega }_{n},{\omega }_{n}^{2},\cdots ,{\omega }_{n}^{n-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{n}=Circ\left(0,1,0,\cdots ,0\right)$

1) ${U}_{n}^{-1}{J}_{n}{U}_{n}={D}_{n}$

2) 若 $C=Circ\left({a}_{0},{a}_{1},\cdots ,{a}_{n-1}\right)$${P}_{v}={a}_{0}+{a}_{1}X+\cdots +{a}_{n-1}{X}^{n-1}$ ，则

${U}_{n}^{-1}C{U}_{n}=diag\left({P}_{v}\left(1\right),{P}_{v}\left({\omega }_{n}\right),\cdots ,{P}_{v}\left({\omega }_{n}^{n-1}\right)\right)$

3. 矩阵的构造及例子

$C={I}_{n}\otimes {C}_{0}+{J}_{n}\otimes {C}_{1}+\cdots +{J}_{n}^{n-1}\otimes {C}_{n-1}$ .

${U}^{-1}CU={I}_{n}\otimes {U}_{n}^{-1}{C}_{0}{U}_{n}+{D}_{n}\otimes {U}_{n}^{-1}{C}_{1}{U}_{n}+\cdots +{D}_{n}^{n-1}\otimes {U}_{n}^{-1}{C}_{n-1}{U}_{n}$ .

${V}_{k}={U}_{n}^{-1}{C}_{k}{U}_{n}=diag\left({P}_{k}\left(1\right),{P}_{k}\left({w}_{n}\right),{P}_{k}\left({w}_{n}^{2}\right),\cdots ,{P}_{k}\left({w}_{n}^{n-1}\right)\right)$ ,

${W}_{k}={V}_{0}+{\omega }_{n}^{k}{V}_{1}+{\omega }_{n}^{2k}{V}_{2}+\cdots +{\omega }_{n}^{k\left(n-1\right)}{V}_{n-1}$ .

${U}^{-1}CU=diag\left({W}_{0},{W}_{1},\cdots ,{W}_{n-1}\right)$ 。易知对任意 $0\le k\le n-1,1\le j\le n$

${\left({W}_{k}\right)}_{j,j}=\left(1,{\omega }_{n}^{k},{\omega }_{n}^{2k},\cdots ,{\omega }_{n}^{k\left(n-1\right)}\right)A\left(1,{\omega }_{n}^{j-1},{\omega }_{n}^{2\left(j-1\right)},\cdots ,{\omega }_{n}^{\left(j-1\right)\left(n-1\right)}\right)$

$\forall 0\le k\le n-1,\text{\hspace{0.17em}}\forall 1\le j\le n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}|{\left({W}_{k}\right)}_{j,j}|=n$ .

${F}_{3}^{*}D=\left(\begin{array}{ccc}a& b& c\\ a& b\omega & c{\omega }^{2}\\ a& b{\omega }^{2}& c\omega \end{array}\right)$

1) 若V是对角矩阵，则 $\text{BCCB}\left(V{F}_{n}\right)$$\text{BCCB}\left({F}_{n}V\right)$ 是哈达码矩阵。

2) 若 ${U}_{n}V{U}_{n}^{*}$ 中元素模全为1，则 $\text{BCCB}\left(V\right)$ 是哈达码矩阵。

3) 若 ${F}_{n}V=V{F}_{n}$ ，则 $\text{BCCB}\left(V\right)$ 是哈达码矩阵。

1) 因为 ${U}_{n}$ 是酉矩阵，故  。因此 ${U}_{n}V{F}_{n}{U}_{n}$ 中元素模全为1。由定理3.1知 $\text{BCCB}\left(V{F}_{n}\right)$ 是哈达码矩阵。同理可证 $\text{BCCB}\left({F}_{n}V\right)$ 也是哈达码矩阵。

2) 首先， ${U}_{n}V{U}_{n}={U}_{n}V{U}_{n}^{*}{U}_{n}^{2}={U}_{n}V{U}_{n}^{*}P$ 。因为 ${U}_{n}V{U}_{n}^{*}$ 中元素模全为1，故 ${U}_{n}V{U}_{n}$ 中元素模全为1。由定理3.1知 $\text{BCCB}\left(V\right)$ 是哈达码矩阵。

3) 由 ${F}_{n}V=V{F}_{n}$${U}_{n}V=V{U}_{n}$ ，于是有 ${U}_{n}V{U}_{n}=V{U}_{n}^{2}=VP$ ，故 ${U}_{n}V{U}_{n}$ 中元素模全为1。由定理3.1知 $\text{BCCB}\left(V\right)$ 是哈达码矩阵。□

${R}_{4}=\frac{1}{2}\left(\begin{array}{cccc}1& \sqrt{2}& 1& 0\\ 1& 0& -1& \sqrt{2}\\ -1& \sqrt{2}& -1& 0\\ 1& 0& -1& -\sqrt{2}\end{array}\right)$

$A={R}_{4}{D}_{4}{R}_{4}^{*}=\frac{1}{4}\left(\begin{array}{cccc}2b+a+c& a-c& 2b-\left(a+c\right)& a-c\\ a-c& a+c+2d& c-a& a+c-2d\\ 2b-\left(a+c\right)& c-a& 2b+a+c& c-a\\ a-c& a+c-2d& c-a& a+c+2d\end{array}\right)$

$\left\{\begin{array}{l}|a+c±2b|=4\\ |a+c±2d|=4\\ |a-c|=4\end{array}$ (1)

$A=\left(\begin{array}{cccc}{\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}\\ {\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{3}}\\ {\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{1}}\\ {\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{3}}\end{array}\right)$

$\left\{\begin{array}{l}|a+c+2b|=4\\ |a+c-2b|=4\\ |a-c|=4\end{array}$ (2)

i) 假设 $b=d$ 。由方程组(2)知，存在 $\left({\theta }_{1},{\theta }_{2},{\theta }_{3}\right)\in {ℝ}^{3}$ 使得

$\left\{\begin{array}{l}a+c+2b=4{\text{e}}^{i{\theta }_{1}}\\ a+c-2b=4{\text{e}}^{i{\theta }_{2}}\\ a-c=4{\text{e}}^{i{\theta }_{3}}\end{array}$

$A=\left(\begin{array}{cccc}{\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{3}}& {\text{e}}^{i{\theta }_{2}}\\ {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{3}}\\ {\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{2}}\\ {\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}\end{array}\right)$

ii) 假设 $b=-d$ 。同理，对任意 $\left({\theta }_{1},{\theta }_{2},{\theta }_{3}\right)\in {ℝ}^{3}$ ，若取

$A=\left(\begin{array}{cccc}{\text{e}}^{i{\theta }_{1}}& {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{3}}& {\text{e}}^{i{\theta }_{2}}\\ {\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}\\ {\text{e}}^{i{\theta }_{3}}& -{\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{2}}\\ {\text{e}}^{i{\theta }_{2}}& {\text{e}}^{i{\theta }_{1}}& -{\text{e}}^{i{\theta }_{2}}& -{\text{e}}^{i{\theta }_{3}}\end{array}\right)$

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