Estimation the system (A) for known input and output [Y = A * X]

Hi All,

For the case of having known input and known output, is there any way to estimate the matrix A in which y = A * x?

I have input matrix (n X 2) which is the position of the object in 2d-Screen at time t. I also know the out put vector which always contain two values (2 X 1).

I would like to estimate the matrix A in which x (variable size) multiplies by A becomes Y. (Y = A * X).

Is there any way to estimate the matrix A? Is this the Jacobian matrix? or should I follow the system identification?!

I would appreciate if you give me any lead in this.

Thanks

Answer 1

[Cedric]

Unless your are in a special situation, there is an infinity of suitable A's .. and there is a dimension mismatch in your statement. Assuming that X is really nx2 and Y is really 2x1, you must have Y = X.'*A, with A a nx1 vector in fact. You must hence solve:

 y1 = x11 a1 + x12 a2 + .. + x1n an
 y2 = x21 a1 + x22 a2 + .. + x2n an

which is a system of 2 equations with n unknowns ( ai). Unless you are in a special situation, you can therefore define n-2 ai and solve for the two remaining ones. This would provide one particular A that fits the system above. An "interesting" choice for the n-2 ai is to set ai=0 for i>2. This reduces the system above to:

 y1 = x11 a1 + x12 a2 
 y2 = x21 a1 + x22 a2

or simply Y = Xred.'*Ared, with Xred=X(1:2,:) and Ared is what you are looking for. This is a linear system that you can easily solve.

Play a bit with the following to see what happens..

 X = [1 2 3; 5 3 4].' ;
 Y = [7 -3].' ;
 n = size(X, 1) ;

 X1 = X(1:2,:) ;
 A1 = X1.' \ Y ;
 A2 = zeros(n-2, 1) ;
 A =[A1; A2]

 Y-X.'*A

I let you think more in depth for special cases.

Cheers,

Cedric

Answer 2

[Youssef Khmou]

Hi Payam,

i am afraid A is not matrix but vector :

lets try :

 n=10;
 X=rand(n,2);
 Y=rand(2,1);
 A=Y'/X;
 Error=(A*X)'-Y;
 Error(Error<1e-10)=0;

[Jan]

It will have strange side-effects, if you shadow the important builtin function error() by a variable.

[Youssef Khmou]

ok Jan, thanks for reminding .