.. _inColSpace:

Exact Solution or Approximation?
--------------------------------

We know that the solution to an over-determined system is exact when
:math:`\bm{b}` is in the column space of :math:`\bf{A}`, but is an
approximation when it is not. The quickest way to determine if the
solution is exact or an approximation is to test if :math:`\bm{b}` is
independent or a linear combination of the columns of :math:`\bf{A}`.

   Vector :math:`\bm{b}` is in the column space of :math:`\bf{A}` when
   the ranks of :math:`\bf{A}` and the augmented matrix
   :math:`[\mathbf{A}\; \bm{b}]` are the same, which means that
   :math:`\bm{b}` is a linear combination of the columns of
   :math:`\bf{A}`.

MATLAB can help us see this. First, let us consider the case where
:math:`\bm{b}` is in the column space of :math:`\bf{A}` and an exact
solution exists. Notice that ``rank(A)`` is equal to ``rank([A b])``.

.. index:: rank

::

   >> f = @(t) 5 - 2.*t;
   >> t = 1:5;
   >> A = ones(5,2);
   >> A(:,2) = t';
   >> A
   A =
       1     1
       1     2
       1     3
       1     4
       1     5 
   >> b = f(t)';
   >> rank(A)
   ans =
       2
   >> rank([A b])
   ans =
       2
   >> x = A\b
   x =
       5.0000
      -2.0000

Now, let’s add some random noise to the :math:`\bm{b}` vector so that it
is not in the column space of :math:`\bf{A}` and the solution is an
approximation. Notice that ``rank(A)`` is less than ``rank([A b])``.

::

   >> b = b+randn(1,5)';
   >> rank(A)
   ans =
       2 
   >> rank([A b])
   ans =
       3 
   >> x = A\b
   x =
       5.6816
      -2.1410