Finding a basis for the null space of a matrix in reduced row echelon form I'm currently reading this paper (for research) on solving the Lights Out game:  https://www.math.ksu.edu/math551/math551a.f06/lights_out.pdf.  
At the bottom of page 301, they find the basis for the null space of a matrix, just by analyzing the last two columns of the reduced row echelon form.  (Note that the entries are in modulo 2) I have very little experience with linear algebra, so I am unsure how they came up with this basis.  I observe that the bases sorta look like the last two columns until halfway, but I am unsure if that actually means anything.  Can someone guide me through the process?
 A: 
Prove $\mathbf{n_1}$ belongs to the nullspace of $E$.

It suffices to prove $\mathbf{En_1}=0$ or $\langle \mathbf{n_1},\mathbf{r_i} \rangle=0$ for any row vector $\mathbf{r_i}$ of $E$.
Note that $\mathbf{n}_1$ is not entirely the same as the second last column of $E$, in which $(\mathbf{n}_1)_{24,1}=1,(\mathbf{n}_1)_{25,1}=0$ and the rest is the same.
For $1 \le i \le 23$, any $i$-th row vector $\mathbf{r_i}$ of $E$ has $(\mathbf{r_i})_{i,1}=1$ and $(\mathbf{r_i})_{j,1}=0$ for $1 \le j \le 23, j \ne i$.We deduce $$\langle \mathbf{n_1},\mathbf{r_i} \rangle = (\mathbf{r_i})_{i,1} (\mathbf{n_1})_{i,1}+ (\mathbf{r_i})_{24,1}(\mathbf{n_1})_{24,1}=(\mathbf{n_1})_{i,1}+(\mathbf{r_i})_{24,1}.$$ Note that $(\mathbf{n_1})_{i,1}=(\mathbf{r_i})_{24,1}$ so $\langle \mathbf{n_1},\mathbf{r_i} \rangle=0$.
For $24 \le i \le 25$, then $\mathbf{r_i}=0$ so of course $\langle \mathbf{n_1},\mathbf{r_i} \rangle=0$.
Similarly, we can prove that $\mathbf{n_2}$ belongs to the nullspace of $E$.

Prove $\mathbf{n_1},\mathbf{n_2}$ are linearly independent.

This is obviously true by simply looking at the $24$-th and $25$-th entries of two vectors.

$\mathbf{n_1},\mathbf{n_2}$ is a basis of nullspace of $E$.

We know $\mathbf{n_1},\mathbf{n_2}$ are linearly independent in $\text{Null}(E)$, and note that $\text{dim } \text{Null}(E)=2$ so we are done.
