How to find the NULL SPACE of the matrix over Finite Field of size 2. A.x=0
For example, I have the following matrix. I know how to convert into reduced echolean form. But then how can I find the NULL SPACE matrix from the reduced echolean form. OR Is there any other method apart from reduced echolean form to find the null space of the matrix ?
Matrix A Reduce echolean form(using exlusive(^) operation between rows)
1 1 1 0 1 1 1 0
1 1 0 1 -> 0 1 0 1
1 0 1 1 0 0 1 1
0 1 1 1 0 0 0 1
9 * 9 -> Reduce Row echolean form.
1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0
1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
1 0 0 1 1 0 1 1 0 0 0 0 1 0 0 0 0 0
0 1 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0
0 0 1 0 1 1 0 1 1 0 0 0 0 0 1 0 0 0
0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1
I have to program it in java that I will manage but tell how to find the null space matrix from the reduce echolean form.
For this particular matrix, row reducing it has just shown that the rank of the matrix is $4$, hence by the Rank-Nullity Theorem, the nullity is $0$. Hence the null space is $\{(0,0,0,0\}$. But in general...
A brute force way to find the null space of $A_{n \times n}$ is to check all $2^n$ vectors $\mathbf{x}$ as to whether or not $A\mathbf{x}=\mathbf{0}$. The inefficiency might not be a big deal for small $n$ (especially if we do it on a computer).
In most cases, it'll be more efficient to find the null space of $A$ from its row echelon form $R$ (or maybe even easier would be reduced row echelon form), then we solve $R\mathbf{x}=\mathbf{0}$ via back substitution. The columns without a leading entry correspond to variables which can be considered free, and the remaining variables can be solved in terms of the free variables.
To convert the matrix to reduced row echelon form, we can perform the following row operations: \begin{align*} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_2 \gets R_2+R_1} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_3 \gets R_3+R_1} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_4 \gets R_4+R_2} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} \\ \xrightarrow{R_4 \gets R_4+R_3} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_1 \gets R_1+R_2} \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_1 \gets R_1+R_3} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_2 \gets R_2+R_4} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \xrightarrow{R_3 \gets R_3+R_4} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \end{align*}
However, since the rank of the matrix is $4$, we know its invertible, and thus will have reduced row echelon form equal to the identity matrix $I$.