在Matlab中求解矩阵方程 [英] Solve matrix equation in matlab

问题描述

我有一个c = Ax + By类型的方程，其中c，x和y是维数为50,000 X 1的向量，而A和B是维数为50,000 X 50,000的矩阵.

I have an equation of the type c = Ax + By where c, x and y are vectors of dimensions say 50,000 X 1, and A and B are matrices with dimensions 50,000 X 50,000.

在已知c，x和y的情况下，Matlab中是否可以找到矩阵A和B?

Is there any way in Matlab to find matrices A and B when c, x and y are known?

我有大约100,000个c，x和y样本. A和B都保持不变.

I have about 100,000 samples of c, x, and y. A and B remain the same for all.

推荐答案

让X为您获得的所有100,000个x的集合(这样，X的第i列等于x_i -th向量).
以相同的方式，我们可以将Y和C分别定义为y s和c s的2D集合.

Let X be the collection of all 100,000 xs you got (such that the i-th column of X equals the x_i-th vector).
In the same manner we can define Y and C as 2D collections of ys and cs respectively.

您要解决的是A和B这样的

C = AX + BY

您有2 * 50,000 ^ 2个未知数(A和B的所有条目)和numel(C)方程.

You have 2 * 50,000^2 unknowns (all entries of A and B) and numel(C) equations.

因此，如果您拥有的数据向量数量为100,000，则您只有一个解(最多取决于线性样本).如果样本数量超过100,000，则可以寻求最小二乘解.

So, if the number of data vectors you have is 100,000 you have a single solution (up to linearly dependent samples). If you have more than 100,000 samples you may seek for a least-squares solution.

重写:

C = [A B] * [X ; Y]  ==>  [X' Y'] * [A';B'] = C'

所以，我想

[A' ; B'] = pinv( [X' Y'] ) * C'

在matlab中:

ABt = pinv( [X' Y'] ) * C';
A = ABt(1:50000,:)';
B = ABt(50001:end,:)';

如果我错了，请纠正我...

Correct me if I'm wrong...

似乎在这里围绕维数大惊小怪.因此，我将尝试使其尽可能清晰.

It seems like there is quite a fuss around dimensionality here. So, I'll try and make it as clear as possible.

模型:有两个(未知)矩阵A和B，每个矩阵的大小为50,000x50,000(总共5e9个未知数).
观察结果是个向量的三元组:(x，y，c)每个这样的向量都有50,000个元素(每个样本 总共有150,000个观察点) ).模型的基本假设是，在此模型中，c = Ax + By生成了一个观测值. 任务:给出了n个观测值(即个向量的n个三元组 {(x_i，y_i，c_i)} _ i=1..n )的任务是发现A和B.

Model: There are two (unknown) matrices A and B, each of size 50,000x50,000 (total 5e9 unknowns).
An observation is a triplet of vectors: (x,y,c) each such vector has 50,000 elements (total of 150,000 observed points at each sample). The underlying model assumption is that an observation is generated by c = Ax + By in this model.
The task: given n observations (that is n triplets of vectors { (x_i, y_i, c_i) }_i=1..n) the task is to uncover A and B.

现在，每个样本(x_i，y_i，c_i)在未知的A和B中产生50,000个形式为c_i = Ax_i + By_i的方程. 如果样本n的数量大于超过100,000，则表示方程组超过50,000 * 100,000(> 5e9)，并且系统超出约束.

Now, each sample (x_i,y_i,c_i) induces 50,000 equations of the form c_i = Ax_i + By_i in the unknown A and B. If the number of samples n is greater than 100,000, then there are more than 50,000 * 100,000 ( > 5e9 ) equations and the system is over constraint.

要以矩阵形式编写系统，我建议将所有观察结果堆叠到矩阵中:

To write the system in a matrix form I proposed to stack all observations into matrices:

• 第i列的大小为50,000 x n的矩阵X等于观察到的x_i
• 第i列的大小为50,000 x n的矩阵Y等于观察到的y_i
• 第i列的大小为50,000 x n的矩阵C等于观察到的c_i

• A matrix X of size 50,000 x n with its i-th column equals to observed x_i
• A matrix Y of size 50,000 x n with its i-th column equals to observed y_i
• A matrix C of size 50,000 x n with its i-th column equals to observed c_i

有了这些矩阵，我们可以将模型编写为:

With these matrices we can write the model as:

C = A * X + B * Y

C = A*X + B*Y

我希望这可以使事情变得清晰起来.

I hope this clears things up a bit.

感谢@Dan和@woodchips对您的关注和启发性的评论.

Thank you @Dan and @woodchips for your interest and enlightening comments.

编辑(2):
将以下代码提交到八度.在此示例中，我只使用2个而不是50,000个维，而不是我为n=100确定的n=100,000个观测值:

EDIT (2):
Submitting the following code to octave. In this example instead of 50,000 dimension I work with only 2, instead of n=100,000 observations I settled for n=100:

n = 100;
A = rand(2,2);
B = rand(2,2);
X = rand(2,n);
Y = rand(2,n);
C = A*X + B*Y + .001*randn(size(X)); % adding noise to observations 
ABt = pinv( [ X' Y'] ) * C';

检查地面真实模型(A和B)与恢复的ABt之间的差异:

Checking the difference between ground truth model (A and B) and recovered ABt:

ABt - [A' ; B']

收益

  ans =

   5.8457e-05   3.0483e-04
   1.1023e-04   6.1842e-05
  -1.2277e-04  -3.2866e-04
  -3.1930e-05  -5.2149e-05

哪个足够接近零. (请记住，观察结果很嘈杂，解决方案是最小二乘法).

Which is close enough to zero. (remember, the observations were noisy and solution is a least-square one).

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