Solve Matrix Equation in Matrix

Let be the X

collection of all 100,000 X

you got (so the i

th column X

equals the x_i

th vector).
In the same way we can define Y

and C

as two-dimensional collections Y

and C

respectively.

What you want to solve is A

and B

so that

C = AX + BY

      

        
        
        
      

    

You have 2 * 50,000 ^ 2 unknowns (all equation records A

and B

) and numel(C)

.

So, if you have 100,000 data vectors, you have one solution (up to linearly dependent samples). If you have more than 100,000 samples, you can search for a least squares solution.

rewriting:

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

      

        
        
        
      

    

So I suppose

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

      

        
        
        
      

    

In Matlab:

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

      

        
        
        
      

    

Correct me if I'm wrong ...

EDIT:
It looks like there is quite a bit of fuss around the dimension here. So I'll try to make it as clear as possible.

Model: There are two (unknown) matrices A

and B

, each of which is 50,000 x 50,000 in size (5e9 unknowns in total).
Observation is a triplet vectors ( X

, Y

, C

), each such vector has elements 50000 (total 150000 points observed in each sample). The underlying assumption of the model is that c = Ax + By

observation is created in this model .
Objective: taking into account n

the observations (i.e., n

triplets {vectors ( x_i

, y_i

, c_i

)} _ i=1..n

) task is to open A

and B

.

Now, each sample ( x_i

, y_i

, c_i

) induces 50,000 of equations c_i = Ax_i + By_i

in the unknown A

and B

. If the number of samples is n

more than 100,000, then there are more than 50,000 * 100,000 (> 5e9) equations, and the system is limited .

To write the system in matrix form, I suggested collecting all observations in matrices:

• A X
  
  50,000x matrix n
  
  with its i
  
  th column is equal to the observedx_i
  
  
• A Y
  
  50,000x matrix n
  
  with its i
  
  th column is equal to the observedy_i
  
  
• A C
  
  50,000x matrix n
  
  with its i
  
  th column is equal to the observedc_i
  
  

With these matrices, we can write the model as:

C = A * X + B * Y

Hope this makes things a little easier.

Thanks @Dan and @woodchips for your interest and enlightening comments.

EDIT (2):
By sending the following code to octave . In this example, instead of 50,000 measurements, I only work with 2, instead of the n=100,000

observations I set 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';

      

        
        
        
      

    

Check the difference between the ground truth model ( A

and B

) and reduced ABt

:

ABt - [A' ; B']

      

        
        
        
      

    

Yield

  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

      

        
        
        
      

    

This is close enough to zero. (remember observations were noisy and the solution was least square).