Why matrix division returns different answers?

Hello fellow Matlabers,

So I have these two matrices:

A = [0.13 -0.022; 0.064 0.006; 0.132 -0.03; 0.012 -0.028];
b = [-0.0287 0.01355];

I want to solve for x where Ax = b. I tried different ways to compute x.

1- Left matrix division

x = A\b

x =

         0
         0
   -0.1921
   -0.2781

x = mldivide(A,b)

x =

         0
         0
   -0.1921
   -0.2781

2- QR decomposition

[Q,R] = qr(A);
x = R\Q.'*b % which used matrix left division!
x =

         0
         0
   -0.1921
   -0.2781

3- LSCOV

x = lscov(A,b)
x =

         0
         0
   -0.1921
   -0.2781   

4- Least squares (extra)

x = lsqr(A,b)
lsqr converged at iteration 2 to a solution with relative residual 4.7e-16.

x =

   -0.0898
    0.0969
   -0.1554
   -0.2261

5- Optimization (extra)

function f = objFun(x, A, b)
    f = norm(b - A*x);
end  

x0 = [0.25 0.25 0.25 0.25]';
options = optimset('TolFun', 1e-15, 'TolX', 1e-15);
[x,fval,exitflag,output] = fmincon(@(x)objFun(x, A, b), x0, [],[],[],[],[],[],[],options)

x =

   -0.1216
    0.2686
   -0.2192
   -0.0960

Problem

I am trying to understand what happens behind the left division (step by step). When I simply do this, the answer is different:

x = pinv(A)*b

ans =

   -0.0898
    0.0969
   -0.1554
   -0.2261

Why isn't the answer the same as what I got for methods 1, 2, and 3? I need a step by step implementation of the matrix left division (without using the left division in the implementation again). I could not find any documentation about the implementation of this division. I really don't know how to arrive at the same answer manually. Any help would be much appriciated.

Thank you!

[Stephen23]

" I could not find any documentation about the implementation of this division"

https://www.mathworks.com/help/matlab/ref/mldivide.html#bt4jslc-6

https://www.mathworks.com/help/matlab/math/systems-of-linear-equations.html

Answer 1

[Bruno Luong]

When you have a underdetermined linear system, meaning you "solve"

A*x = b

where A has non-zero kernel, or in other word

null(A)

returns non-empty set of vectors, you have many choice of x that gives the same map

b = A*x

The A\b use QR with permutation (the "permutation" matters in other example different thean yours ) to return the so called "basis solution", meaning rhe solution has most 0s, the non-zeros terms corresponds to independent columns of A, selected by QR decomposition with permutation.

The pinv however gives the least 2-norm solution, meaning the solution that minimizes norm(x,2) among all solutions that gives the same A*x. It's n afine space of dimension numvber of column of null(A)

If your system is overdetermined, meaning rank(A) < size(A,2), then the solution of

A*x = b

may not be reached, in this case the above equation is approximated by

x = argmin norm(A*x-b,2)

and this holds for A\b or pinv(A)*b.

NOTE that the linear sysyem of equations can be underdetermined and overdetermned at the same time.

Answer 2

[Ahmad AlAttar]

[Q,R,P] = qr(A); % QRP decomposition

rearrange_A = A*P; % rearrange matrix A

basis_A = rearrange_A(:,1:2); % extract left half of matrix

x = pinv(basis_A)*b;

[Bruno Luong]

How does this snip of code answers your question? It returns the result that is not even compatible with

A*x = b

with A that is fat like 2 x 4 in size.