I am trying to find the null space of a matrix mod 2. So far I have tried to implement basic Gaussian elimination.
Something happened that should've been very easy to solve but it's late and I can't think of a solution. Here's an example I made up:
$$\left( \begin{array}{ccccc} 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 \end{array} \right)$$
I've finished reducing the first column. I want to reduce the second column but there's a $0$ there.
In the end I am trying to find linear dependencies. Everything I've searched up has been for single solutions.
Edit: Apparently these are called free or pivot columns. If there were something in the second column below the zero it would be swapped.
EDIT: I initially made a stupid error in my conclusion. It has been fixed now - the null space is 2D!
Forgive me, if I am too stupid to understand what you are asking! But in your example $$ \begin{pmatrix} 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 \end{pmatrix} $$ I get the reduced row-echelon form to be $$ \begin{pmatrix}1&1&0&0&0\\0&0&1&0&0\\0&0&0&1&1\\0&0&0&0&0\end{pmatrix} $$ The above corresponds to the equations (noting that we are calculating mod 2) $$ \begin{align} x_1+x_2&=0&\iff&&x_1=x_2\\ x_3&=0\\ x_4+x_5&=0&\iff&&x_4=x_5 \end{align} $$ and conlude that the nullspace has the form $$ \begin{pmatrix}s\\s\\0\\t\\t\end{pmatrix}= s\begin{pmatrix}1\\1\\0\\0\\0\end{pmatrix}+ t\begin{pmatrix}0\\0\\0\\1\\1\end{pmatrix} $$ where $s,t\in\mathbb B=\{0,1\}$.