I have matrix A, calculate null(A) but then A*null(A) doesn't give me 0?

%a) The traffic flow is an overdetermined system 12 variables to 9 equations
A = [-1 0 0 0 0 0 0 1 0 0 0 0;
    1 1 0 0 0 0 0 0 0 0 -1 0;
    0 -1 1 0 0 0 0 0 0 0 0 0; 
    0 0 -1 -1 0 0 0 0 0 0 0 1;
    0 0 0 1 1 0 0 0 0 0 0 0 ;
    0 0 0 0 1 1 0 0 -1 0 0 0;
    0 0 0 0 0 -1 1 0 0 0 0 0;
    0 0 0 0 0 0 -1 -1 0 1 0 0;
    0 0 0 0 0 0 0 0 1 -1 1 -1];

b = [0; 150; 20; -410; 180; 210; 80; -230; 0];

%b)Gives five free variables, so there is not a unique solution, 
%meaning the traffic flow depends in other intersects
x = linsolve(A,b);

%e)
nullspace = null(A) %Three vectors in the nullspace
nullVector  = A* nullspace(:,1)   %Shouldn't this give me zero vector?

I am wondering if I calculate null(A) and tree null vectors:

    0.0939    0.3560   -0.3875
   -0.1103    0.2839    0.4184
   -0.1103    0.2839    0.4184
   -0.0000    0.0000    0.0000
    0.0000   -0.0000   -0.0000
    0.4610   -0.0825    0.2202
    0.4610   -0.0825    0.2202
    0.0939    0.3560   -0.3875
    0.4610   -0.0825    0.2202
    0.5549    0.2734   -0.1674
   -0.0164    0.6399    0.0309
   -0.1103    0.2839    0.4184

Shouldn't that mean A*nullspaceVector = 0 ?

But nullVector = A* nullspace(:,1) gives me:

nullVector =

   1.0e-15 *

    0.0139
   -0.1180
    0.3053
   -0.3053
   -0.0714
    0.2776
   -0.1665
    0.2220
    0.0971

[Bruno Luong]

1.0e-15 * somenumber is considered as 0 numerically in floating point calculation.

Illustration of the same thing with simpler example

>> x=[3 4];
>> y=null(x)

y =

   -0.8000
    0.6000

>> x*y

ans =

  -4.4409e-16

Answer 1

[Steven Lord]

Note that the description of the output argument on the documentation page for the null function does not include the character 0. It says that Z satisfies two properties, the key one of which for this question is "A*Z has negligible elements."

I would call 1e-15 or 1e-16 negligible relative to the elements in your matrix.