A note on the ranks of 2 × 2 × 2 and 2 × 2 × 2 × 2 tensors

Using rank-1 reduction formula and the vector space spanned by the real rank-1 matrices, we present a different way to show that the maximum possible rank of the 2?×?2?×?2 tensors over the real field is 3. Following, we obtain that the maximum rank of the 2?×?2?×?2?×?2 tensors over the real field is less than or equal to 5 and propose another way to show that the maximum rank of the 2?×?2?×?2?×?2 tensors over the complex field is 4, except one special case.     

求1~2 2~2 3~2 … n~2的一种方法

因为每个三角形数阵各数字之和=12 22 32 … n2．每个三角形数阵共有(1 n)n/2 个数．现将三个三角形数阵平移使其重合,则重合后每个小圆圈内三数和为2n 1,所以重     

利用|z_1 z_2|~2 |z_1-z_2|~2=2|z_1|~2 2|z_2|~2解题

关于复数模的有关性质之一有公式|z_1 z_2|~2 |z_1-z_2|~2=2|z_1|~2 2|z_2|~2其几何意义是:平行四边形两对角线的平方和等于四边平方和,利用它解决一类有关复数模的问题不但有效,而且解题过程简单,方法新颖。例1 已知|z 3 4i|~2 |z-3-4i|~2=80求|z|:并说明z点的轨迹表示的图形。分析若设z=x yi代入已知整理,则会步骤冗长,利用     

On the classification of nonsingular 2×2×2×2 hypercubes

As a first step in the classification of nonsingular 2×2×2×2 hypercubes up to equivalence, we resolve the case where the base field is finite and the hypercubes can be written as a product of two 2×2×2 hypercubes. (Nonsingular hypercubes were introduced by D. Knuth in the context of semifields. Where semifields are related to hypercubes of dimension 3, this paper considers the next case, i.e., hypercubes of dimension 4.) We define the notion of ij-rank (with 1 ≤ i < j ≤ 4) and prove that a hypercube is the product of two 2×2×2 hypercubes if and only if its 12-rank is at most 2. We derive a ‘standard form’ for nonsingular 2×2×2×2 hypercubes of 12-rank less than 4 as a first step in the classification of such hypercubes up to equivalence. Our main result states that the equivalence class of a nonsingular 2×2×2×2 hypercube M of 12-rank 2 depends only on the value of an invariant δ ₀(M) which derives in a natural way from the Cayley hyperdeterminant det₀ M and another polynomial invariant det M of degree 4. As a corollary we prove that the number of equivalence classes is (q + 1)/2 or q/2 depending on whether the order q of the field is odd or even.     

用((a b)/2)~2≤(a~2 b~2)/2解题两例

对于a、b∈R,易知(a b/2)2≤a2 b2/2恒成立,此不等式反映了任意两个实数的和与这两个数的平方和的大小关系,巧用这一不等式可以妙解一些数学不等关系的问题．请看以下两例: 1．比较大小例1试比较2~(1/2) 3~(1/2) 4~(1/2)与30~(1/2)的大小．解∵(a b/2)2≤a2 b2/2,     

The rank of a 2 × 2 × 2 tensor

As computing power increases, many more problems in engineering and data analysis involve computation with tensors, or multi-way data arrays. Most applications involve computing a decomposition of a tensor into a linear combination of rank-1 tensors. Ideally, the decomposition involves a minimal number of terms, i.e. computation of the rank of the tensor. Tensor rank is not a straight-forward extension of matrix rank. A constructive proof based on an eigenvalue criterion is provided that shows when a 2?×?2?×?2 tensor over ? is rank-3 and when it is rank-2. The results are extended to show that n?×?n?×?2 tensors over ? have maximum possible rank n?+?k where k is the number of complex conjugate eigenvalue pairs of the matrices forming the two faces of the tensor cube.     

商高定理与方程x~2+y~2+z~2=w~2

华罗庚著《数論导引》中“商高定理”一节,見有方程 x~2+y~2+z~2=w~2 (1)习題一则,遂默思其解,得到了解法数种。現在写出来向同志們請教。 (一) 我們称方程 x~2+y~2=z~2 (2)的解[x,y,z]为“商高数”。如有两組商高教,其一組之第三項(或其倍数)适与另一組之第一或第二項(或其倍数)相等,以第一組之前两項,代另一組之前两項中之一項,那么,就得到方程(1)的一組解。设两組商高数:     

Canonical forms of 2 × 2 × 2 and 2 × 2 × 2 × 2 arrays over 𝔽2 and 𝔽3

We consider arrays of size 2?×?2?×?2 and 2?×?2?×?2?×?2 over the fields with two and three elements. We use computer algebra to determine the canonical forms of these arrays with respect to the action of the semidirect product of general linear groups with the symmetric group. For each canonical form, we determine the size of its orbit and the rank of the arrays in its orbit.     

2×2矩阵的平方根

２×２矩阵的平方根［美］ＤＯＮＡＬＤＳＵＬＬＶＡＮ在Ｍａｃｋｉｎｎｏｎ最近的论文［１］里，叙述了求２×２矩阵平方根的四种方法．这些方法的第一个方法要求那些求平方根的矩阵是可以对角化的．后来，这个方法被Ｓｃｏｔ用来求２×２矩阵的全部平方根［２］．一个奇...     

2×2 Hermite矩阵几何

设D是有对合a→的除环, 假设是D的真子域并且含在D的中心域中. 指出: 如果D的特征不等于2, 则D是F上可离二次扩域或者是F上广义四元数除环; 如果D的特征等于2, 则D是F上可离二次扩域, 因此迹映射Tr:D→F,恒为满射,从而已有的 Hermite矩阵几何的基本定理中关于D的假设(ii)可以删去. 万哲先已经证明了当D为域时2×2 Hermite矩阵几何的基本定理. 继续证明了当D为特征不等于2的广义四元数除环时,D上2×2 Hermite矩阵几何的基本定理.     

2×2矩阵的平方根

在线性代数中我们能够用适当的公式计算矩阵的逆,特征值,特征向量。这篇短文的目的是给出一个2×2矩阵平方根的简单公式。作为卡莱——哈密顿定理的一个应用。     

用(a~2 b~2)(c~2 d~2)≥(ac bd)~2解三角题

高中《代数》（必修）下册第１５页第６题可改述为：（ａ２＋ｂ２）（ｃ２＋ｄ２）≥（ａｃ＋ｂｄ）２○＊当且仅当ａ／ｃ＝ｂ／ｄ时取等号．灵活巧妙地运用○＊式，可使某些三角问题简捷获解．例１已知Ａ，Ｂ都是锐角，且ｃｏｓＡ＋ｃｏｓＢ－ｃｏｓ（Ａ＋Ｂ）＝３／２，...     

导出1~2 2~2 3~2 … n~2的公式的一种方法

在贵刊2001年2月上期及2001年12月上期上,分别刊登了谢秀英与张立华二位老师的关于导出12 22 32 … n2(设为Sn)的公式的几种方法,很受启发.笔者在这一部分的教学过程中,与学生一起发现了12 22 32 … n2的公式导出的另外一种方法. 看图1的数阵:可以看作是一个“等边三角形”,其共有n行,每行的数字相同(从上至下每一行的数都是这一行的行数,且每一行数的个数与行数也相同).不难看出每行所有数的和分     

a~2+b~2≥2ab的一点引伸

众所周知,不等式a~2+b~2≥2ab是一个应用广泛的重要不等式,由此容易推出以下两个不等式: a~2+b~2+c~2≥ab+bc+ca; a~2+b~2+c~2+d~2≥ab+bc+cd+da。进一步推广可得更一般的如下: 定理当a_1∈R (i=1,2,…,n)时,     

Pell方程组x~2-(c~2-1)y~2=y~2-2p_1p_2p_3z~2=1的公解

设p_1,p_2,p_3为不同的奇素数,c1是整数.给出了Pell方程组x~2-(c~2-1)y~2=y~2-2p_1p_2p_3z~2=1的所有非负整数解(x,y,z),从而推广了Keskin (2017)和Cipu(2018)等人的结果.     

环F_2+uF_2+u~2F_2上循环码的Gray距离

定义了环R=F2+uF2+u2 F2(u3=0)到F32的一个新的Gray映射.首先介绍环R上奇长度的循环码的挠码,给出了各阶挠码的生成多项式.利用一阶挠码与二阶挠码确立了R上奇长度的循环码的Gray距离.