Matlab:如何将矩阵转换为Toeplitz矩阵 [英] Matlab: How to convert a matrix into a Toeplitz matrix

问题描述

考虑离散动力系统，其中x[0]=rand()表示系统的初始条件.

Considering a discrete dynamical system where x[0]=rand() denotes the initial condition of the system.

我通过以下步骤生成了一个m×n矩阵-生成m个具有m个不同初始条件的向量，每个初始条件的维数为N(N表示样本或元素的数量).该矩阵称为R.如何使用R创建Toeplitz矩阵T? Ť 在数学上，

I have generated an m by n matrix by the following step -- generate m vectors with m different initial conditions each with dimension N (N indicates the number of samples or elements). This matrix is called R. Using R how do I create a Toeplitz matrix, T? T Mathematically,

 R =               [ x_0[0], ....,x_0[n-1];
                    ...,        ,.....;
                    x_m[0],.....,x_m[n-1]]

toeplitz矩阵T =

The toeplitz matrix T =

                        x[n-1], x[n-2],....,x[0];
                        x[0], x[n-1],....,x[1];
                          :    :           :
                         x[m-2],x[m-3]....,x[m-1]

我尝试使用toeplitz(R)，但是尺寸发生了变化.从数学角度看，尺寸应该没有变化.

I tried working with toeplitz(R) but the dimension changes. The dimension should no change, as seen mathematically.

推荐答案

根据提供的论文(Yu等人的用于压缩感测的Toeplitz结构混沌感测矩阵)，有两个混沌感测矩阵涉及.让我们分别探讨它们.

According to the paper provided (Toeplitz structured chaotic sensing matrix for compressive sensing by Yu et al.) there are two Chaotic Sensing Matrices involved. Let's explore them separately.

1. 混沌感测矩阵( A 部分)

很明显，要创建这样的矩阵，您必须使用m个不同的初始条件(范围为[0; 1 [])构建m个独立的信号(序列)，然后每行连接这些信号(即，一个信号=一行).这些信号中的每一个都必须具有长度N.这实际上是您的矩阵R，它已按原样正确评估.尽管我想提出一个代码改进建议:您可以直接在每行中构建这样的矩阵，而不是先构建一列然后转置矩阵:

It is clearly stated that to create such matrix you have to build m independent signals (sequences) with m different initials conditions (in range ]0;1[) and then concatenate such signals per rows (that is, one signal = one row). Each of these signals must have length N. This actually is your matrix R, which is correctly evaluated as it is. Although I'd like to suggest a code improvement: instead of building a column and then transpose the matrix you can directly build such matrix per rows:

R=zeros(m,N);
R(:,1)=rand(m,1); %build the first column with m initial conditions

请注意:通过运行randn()，您选择具有高斯(正态)分布的值，这些值可能不在论文所述的[0; 1]范围内(在等式9的正下方) ).相反，通过使用rand()，您可以在此范围内获取均匀分布的值.

Please note: by running randn() you select values with Gaussian (Normal) distribution, such values might not be in range ]0;1[ as stated in the paper (right below equation 9). As instead by using rand() you take uniformly distributed values in such range.

之后，您可以根据for循环分别构建每一行:

After that, you can build every row separately according to the for-loop:

for i=1:m
    for j=2:N %skip first column
        R(i,j)=4*R(i,j-1)*(1-R(i,j-1));
        R(i,j)=R(i,j)-0.5;
    end
end

1. Toeplitz混沌传感矩阵(部分 B )

在 B 节的开头明确指出，要构建Toeplitz矩阵，您应该考虑具有给定，单个初始条件的单个序列x.因此，让我们构建这样的序列:

It is clearly stated at the beginning of Section B that to build the Toeplitz matrix you should consider a single sequence x with a given, single, initial condition. So let's build such sequence:

x=rand();
for j=2:N %skip first element
   x(j)=4*x(j-1)*(1-x(j-1));  
   x(j)=x(j)-0.5;
end

现在，您可以考虑构建矩阵:

Now, to build the matrix you can consider:

• 第一行的样子如何?好吧，它看起来像序列本身，但是翻转了(即不是从 0 变为 n-1 ，而是从 n-1 到 0 )
• 第一列的外观如何?这是x中最后一个项目，与范围从 0 到 m-2
的元素连接

• how do the first row looks like? Well, it looks like the sequence itself, but flipped (i.e. instead of going from 0 to n-1, it goes from n-1 to 0)
• how do the first column looks like? It is the last item from x concatenated with the elements in range 0 to m-2

然后构建第一行(r)和第一列(c):

Let's then build the first row (r) and the first column (c):

r=fliplr(x);
c=[x(end) x(1:m-1)];

请注意:在Matlab中，索引从1开始，而不是从0开始(因此，不是从 0 到 m-2 ，我们从 1 转到 m-1 ).另外，end表示给定数组中的最后一个元素.

Please note: in Matlab the indices start from 1, not from 0 (so instead of going from 0 to m-2, we go from 1 to m-1). Also end means the last element from a given array.

现在，通过查看toeplitz()函数的帮助，可以清楚地说明可以通过指定第一行和第一列来构建非平方的Toeplitz矩阵.因此，最后，您可以构建如下矩阵:

Now by looking at the help for the toeplitz() function, it is clearly stated that you can build a non-squared Toeplitz matrix by specifying the first row and the first column. Therefore, finally, you can build such matrix as:

T=toeplitz(c,r);

如本文所报道的那样，此类矩阵确实具有尺寸m*N.

Such matrix will indeed have dimensions m*N, as reported in the paper.

即使作者都称他们为\ Phi，但实际上它们是两个单独的矩阵.
他们不采用贝塔矩阵矩阵的托普利兹(托普利茨矩阵不是某种函数或算符)，也不会将贝塔矩阵转化为托普利兹矩阵.
首先，您具有类似Beta的矩阵(即混沌感测矩阵)，然后是Toeplitz- 结构化的混沌感测矩阵:这样的 structure 是Toeplitz矩阵的典型特征，即是对角线常数结构(沿对角线的所有元素都具有相同的值).

Even though the Authors call both of them \P they actually are two separate matrices.
They do not take the Toeplitz of the Beta-Like Matrix (Toeplitz matrix is not a function or operator of some kind), neither do they transform the Beta-Like Matrix into a Toeplitz-matrix.
You have the Beta-Like Matrix (i.e. the Chaotic Sensing Matrix) at first, and then the Toeplitz-structured Chaotic Sensing Matrix: such structure is typical for Toeplitz matrices, that is a diagonal-constant structure (all elements along a diagonal have the same value).

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