lsqr converged but returns large residual

I'm using lsqr. It converges, but I get this message:

 lsqr converged at iteration 10 to a solution with relative residual 0.62.

I read in the documentation that relative residual is less than tolerance if lsqr converge; why do i get this residual? This is my code:

tol=1e-2;
maxit=30;
x=lsqr(A,B,tol,maxit);

Thanks for the help

Answer 1

[John D'Errico]

No. There is absolutely no presumption that lsqr (or ANY such tool) will give an exact solution in terms of essentially a zero residual. Consider this example:

A = rand(3,2)
A =
    0.81472      0.91338
    0.90579      0.63236
    0.12699      0.09754

b = rand(3,1)
b =
     0.2785
    0.54688
    0.95751

x = lsqr(A,b)
lsqr converged at iteration 2 to a solution with relative residual 0.77.
x =
     1.2896
   -0.82073

As you can see, it agrees with that which backslash returns.

A\b
ans =
     1.2896
   -0.82073

Is it exact? OF COURSE NOT!

A*x-b
ans =
   0.022536
    0.10223
    -0.8738

No exact solution exists. This is what you have, a problem with no exact solution.

Merely setting a small tolerance cannot enforce a solution to exist where no solution exists. The tolerance only tells lsqr when to stop iterating. So lets see what you THINK you read in the documentation.

[X,FLAG,RELRES] = lsqr(A,B,...) also returns estimates of the relative
  residual NORM(B-A*X)/NORM(B). If RELRES <= TOL, then X is a
  consistent solution to A*X=B. If FLAG is 0 but RELRES > TOL, then X is
  the least squares solution which minimizes norm(B-A*X).

LSQR returns a relative residual, but as you can read above, there is no presumption that the solution is exact.

[Matteo Cencini]

Thank you very much!

[Vardhil Mehta]

So, how to get lower relative residual?