MATH 372 -- Numerical Linear Algebra

SVD for numerical rank determination

March 26, 2007

As we know, matrices like this A cannot
be represented exactly with any binary or 
decimal floating point number system.

>> A = [1/3 2/3 4/3; 1/6 1/3 2/3; 1/12 1/6 1/3; 1/24 1/12 1/6]

A =

    0.3333    0.6667    1.3333
    0.1667    0.3333    0.6667
    0.0833    0.1667    0.3333
    0.0417    0.0833    0.1667

This means that we are dealing with a perturbed
matrix right from the start, and the perturbation
will most likely change the rank.  For jobs like
computing ranks, this means we must rely on 
more sophisticated methods than just row-reduction.
The SVD gives one good approach.  We first tell
MATLAB to display results in "scientific notation"
with short mantissas:

>> format short e

Then the SVD command (displaying the U, S, V factors)
is:

>> [U,S,V]=svd(A)

U =

  -8.6772e-01   4.9705e-01   1.0530e-17   1.2960e-17
  -4.3386e-01  -7.5741e-01  -4.3644e-01  -2.1822e-01
  -2.1693e-01  -3.7870e-01   8.9830e-01  -5.0851e-02
  -1.0847e-01  -1.8935e-01  -5.0851e-02   9.7457e-01


S =

   1.7604e+00            0            0
            0   3.1032e-17            0
            0            0            0
            0            0            0


V =

  -2.1822e-01  -9.7590e-01            0
  -4.3644e-01   9.7590e-02  -8.9443e-01
  -8.7287e-01   1.9518e-01   4.4721e-01

The singular values are the diagonal entries in S:

    1.7604e+00   3.1032e-17   0

Even though the second is not exactly zero, we suspect
the rank should be 1 since it is so small.  MATLAB agrees:

rank(A)

ans =

     1