Factor $f(x) = x^4+x^2+x+1$ over $F_2$ by using the Berlekamp's algorithm.
Here to factor the polynomial $f(x)$ I did the following steps.
Step 1. $\gcd(f(x), f'(x)) = 1$ and $f(x)$ has no repeated factors.
step 2. compute the powers $x^{2i} \bmod f(x)$ for $0 \leq i \leq 3$. This yields the following congruences:
$$x^0= 1 \bmod f,\ x^2 = x^2 \bmod f,\ x^4 = 1 + x + x^2 \bmod f,\
x^6 = 1 + x + x^3 \bmod f.$$
Step 3. So, the matrix $D$ of order 4 by 4 is given as $$D = \begin{bmatrix}1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\1 & 1 & 1 & 0\\1 &1 & 0 & 1\end{bmatrix}$$ and the matrix $D-I$ is
$$D-I = \begin{bmatrix}0 & 0 & 0 & 0\\0 & -1 & 1 & 0\\1 & 1 & 0 & 0\\1 &1 & 0 & 0\end{bmatrix}.$$
Step 4. Compute the basis of the null space for the matrix.
Here is my confusion. I computed the solution by letting eigenvectors $(x,y,z,w)$ such that $-y+z=0,x+y=0$, and I got a basis vector $\{(1,-1,-1,0), (0,0,0,1)\}$ . But from one textbook I have found the solution as $\{(1,0,0,0),(0,0,1,1)\}.$ Did I make a mistake? Hereafter it is easy. Thanks!