Calculation null space(kernel) of matrix

Question

I'm trying to write a program in which one part is related to calculation of kernel of a matrix. Its output and expected output are different. For example,

$$ A = \begin{pmatrix}1&0&1\\ 1&0&0\\ 0&1&1\\ 0&1&0\\ 0&0&1\\ -1&0&0\\ 0&0&-1\\ 0&-1&1\\ 0&-1&0\\ -1&0&1\end{pmatrix} $$

When I calculate its kernel with some programs they give output different, my program is as well. But, when I try to calculate its kernel some others, like SAGE, give output,

1  0  0  0  0  0  0 -2  2  1
0  1  0  0  0  0  0 -1  1  1
0  0  1  0  0  0  0 -1  2  0
0  0  0  1  0  0  0  0  1  0
0  0  0  0  1  0  0 -1  1  0
0  0  0  0  0  1  0  1 -1 -1
0  0  0  0  0  0  1  1 -1  0

The above one is what I expect as output. What is the point that I may overlook?

@Edit, It's my algorithm,

A.transposeInPlace();
FullPivLU<MatrixXf> lu(A);
MatrixXf A_null_space = lu.kernel();
A_null_space.transposeInPlace();

when I run, it gives

 0.5    0   -1    1    0    0    0    0    0  0.5
-0.5    0   -0    0    1    0    0    0    0 -0.5
 0.5    0   -0    0    0    1    0    0    0 -0.5
 0.5    0   -0    0    0    0    1    0    0  0.5
  -1    0    1    0    0    0    0    1    0   -1
-0.5    0    1    0    0    0    0    0    1 -0.5
-0.5    1   -0    0    0    0    0    0    0  0.5

@Edit 2, I'm really but really confused because both matrix seem right! How come?

Answer 1

It sounds like you may have missed a transpose along with way. Your matrix $A$ is 10 x 3, it has rank 3, and as such it will have a kernel of rank 0 by the Rank Nullity Theorem.

However, if you were looking for the kernel of $A^T$, this would have rank 7, and the basis of that kernel would line up with the seven row vectors you provided.