How can I solve the least square minimization Ax=b when b is unknown?

Hi all,

I'm new in the forum. I've been thinking to this problem for a long time, so any suggestion would be appreciated. I've the following problem:

I have this equation

E p = d, where

               E = (x1, y1, z1; 
                    x2, y2, z2; 
                    x3, y3, z3; 
                    x4, y4, z4)

               p = (a; b; c)

               d = (d; d; d; d)

I want to find p using the least square minimization, but also d is unknown.

Do you have any suggestion?

Thank you

Valentina

[Star Strider]

Tell us more about what created that problem. Maybe that will suggest a solution.

Answer 1

[Roger Stafford]

Whatever the value of d is, the following will give the least squares approximation for E*p-[d;d;d;d] = 0.

   x = [x1;x2;x3;x4];
   y = [y1;y2;y3;y4];
   z = [z1;z2;z3;z4];
   A = [sum(x.^2),sum(x.*y),sum(x.*z);
        sum(x.*y),sum(y.^2),sum(y.*z);
        sum(x.*z),sum(y.*z),sum(z.^2)];
   p = d*(A\[sum(x);sum(y);sum(z)]);

where p = [a;b;c].

Answer 2

[John D'Errico]

I would point out that the solution is simply a multiple of p1, where p1 is given as the solution of:

E*p1 = ones(4,1)

Think of p1 as the solution for d=1. So solve for p1 as

p1 = E\ones(4,1);

Then despite the fact that you don't know d, the solution is simply

syms d
p = d*p1

Fairly trivial. Of course, that leaves the solution defined in terms of d as a parameter. If you then decided on the value of d, it simply scales p1 to give your answer.

If the goal is to solve for d also, then by moving d into the set of unknowns, now we have:

phat = [a;b;c;d]

Ehat = [x1, y1, z1, -1; 
        x2, y2, z2, -1; 
        x3, y3, z3, -1; 
        x4, y4, z4, -1];

(This is essentially what Matt did, but he missed a sign on d when he moved it to the left hand side.) Now you will be solving the homogeneous problem

Ehat*phat = 0

If the Ehat matrix is of less than full rank (i.e., it is singular), then a non-degenerate solution always exists. We can get the solution from the nullspace of Ehat. Thus the columns of null(Ehat) will yield a pair of vectors that spans the solution set.

For example:

E = [1 2 3;4 5 6;7 8 9;10 11 13]
rank(E)
ans =
   3

Ehat = [E,-ones(4,1)]
Ehat =
   1     2     3    -1
   4     5     6    -1
   7     8     9    -1
  10    11    13    -1

rank(Ehat)
ans =
   3

phat = null(Ehat)
phat =
    0.57735
   -0.57735
 3.2752e-15
   -0.57735

In fact, any scalar multiple of this vector is also a solution. So here we have

phat = phat/phat(1)
phat =
            1
           -1
   5.6727e-15
           -1

We can now extract a,b,c,d from the above phat vector.

However, if I change E slightly,

E = [1 2 3;4 5 6;7 8 9;10 11 12];
Ehat = [E,-ones(4,1)];

rank(Ehat)
ans =
     2

Ehat now has rank 2, so the solution takes on the form of any linear combination of the columns of phat.

phat = null(Ehat)
phat =
     -0.50648     -0.27714
      0.78363     -0.22934
     -0.27714      0.50648
      0.22934      0.78363

Answer 3

[Valentina]

Thank you both for your answer.

Roger in your solution how should d be defined?

Because if I don't define d, I will get the error: Undefined function or variable 'd'.

Thank you again.

Valentina

[Roger Stafford]

Valentina, there is absolutely no way of obtaining a least squares solution to your problem without knowing the value of d. What I have given you is simply a formula for computing a least squares solution once you know d.

To ask for a specific solution without knowing d is analogous to asking for the minimum value of a set of numbers without knowing what those numbers are. It is obviously impossible.

Answer 4

[Matt J]

 EE = [E,ones(4,1)];

 [U,S,V]=svd(EE);

 p=V(:,end);