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在matlab中求解矩阵方程

[英]Solve matrix equation in matlab

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. 我有一个类型c = Ax + By的等式,其中cxy是尺寸为50,000 X 1的向量, AB是尺寸为50,000 X 50,000的矩阵。

Is there any way in Matlab to find matrices A and B when c , x and y are known? cxy已知时,Matlab中是否有任何方法可以找到矩阵AB

I have about 100,000 samples of c , x , and y . 我有大约100,000个cxy样本。 A and B remain the same for all. AB对所有人保持不变。

Let X be the collection of all 100,000 x s you got (such that the i -th column of X equals the x_i -th vector). X是你得到的所有100,000 x s的集合(这样X的第i列等于x_i -th向量)。
In the same manner we can define Y and C as 2D collections of y s and c s respectively. 以相同的方式,我们可以将YC分别定义为y s和c s的2D集合。

What you wish to solve is for A and B such that 你想要解决的是AB这样的

C = AX + BY

You have 2 * 50,000^2 unknowns (all entries of A and B ) and numel(C) equations. 您有2 * 50,000 ^ 2个未知数( AB所有条目)和numel(C)方程。

So, if the number of data vectors you have is 100,000 you have a single solution (up to linearly dependent samples). 因此,如果您拥有的数据向量数为100,000,则您只有一个解决方案(最多取决于线性相关的样本)。 If you have more than 100,000 samples you may seek for a least-squares solution. 如果您有超过100,000个样本,您可以寻求最小二乘解。

Re-writing: 重新书写:

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

So, I suppose 所以,我想

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

In matlab: 在matlab中:

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

Correct me if I'm wrong... 如我错了请纠正我...

EDIT: 编辑:
It seems like there is quite a fuss around dimensionality here. 这里似乎有很大的维度。 So, I'll try and make it as clear as possible. 所以,我会尽量让它变得清晰。

Model: There are two (unknown) matrices A and B , each of size 50,000x50,000 (total 5e9 unknowns). 模型:有两个(未知)矩阵AB ,每个矩阵大小为50,000x50,000(总共5e9个未知数)。
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 ). 观察是向量的三元组:( xyc )每个这样的载体具有50,000个元素( 每个样品总共150,000个观察点)。 The underlying model assumption is that an observation is generated by c = Ax + By in this model. 基础模型假设是在该模型中由c = Ax + By生成观测值。
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 . 任务:给定n观测值(即n三元组的向量 {( x_iy_ic_i )} _ i=1..n ),任务是揭示AB

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 . 现在,每个样本( x_iy_ic_i )在未知AB c_i = Ax_i + By_i形式为c_i = Ax_i + By_i 50,000个方程。 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 . 如果样本数n 大于 100,000,则存在超过50,000 * 100,000(> 5e9)个方程,并且系统超出约束

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

  • A matrix X of size 50,000 x n with its i -th column equals to observed x_i 矩阵X的大小为50,000 x n ,其第i列等于观察到的x_i
  • A matrix Y of size 50,000 x n with its i -th column equals to observed y_i 矩阵Y的大小为50,000 x n ,其第i列等于观察到的y_i
  • A matrix C of size 50,000 x n with its i -th column equals to observed c_i 矩阵C的大小为50,000 x n ,其第i列等于观察到的c_i

With these matrices we can write the model as: 使用这些矩阵,我们可以将模型编写为:

C = A*X + B*Y C = A * X + B * Y.

I hope this clears things up a bit. 我希望这可以解决一些问题。

Thank you @Dan and @woodchips for your interest and enlightening comments. 感谢@Dan和@woodchips的兴趣和启发性评论。

EDIT (2): 编辑(2):
Submitting the following code to octave . 提交以下代码到八度 In this example instead of 50,000 dimension I work with only 2, instead of n=100,000 observations I settled for n=100 : 在这个例子而不是50,000维度我只使用2,而不是n=100,000观察我确定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';

Checking the difference between ground truth model ( A and B ) and recovered ABt : 检查地面实况模型( AB )与恢复的ABt之间的差异:

ABt - [A' ; B']

Yields 产量

  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

Which is close enough to zero. 哪个足够接近零。 (remember, the observations were noisy and solution is a least-square one). (记住,观察结果是嘈杂的,解决方案是最不正方形的)。

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