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

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%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

#### 1 Comment

Bruno Luong on 21 Oct 2020
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

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### Answers (1)

Steven Lord on 21 Oct 2020
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.

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