Let $P_3$ be the space of polynomials of degree $\leq 3$ . Find the kernel and the image of the linear map $f(x) \mapsto f(x + 1)−f(x)$ Let $P_3$ be the space of polynomials of degree $\leq 3$ over the field $\mathbb{Z}/2\mathbb{Z}$. Find the kernel and the image (that
is, give bases of these spaces) of the linear map $f(x) \mapsto f(x + 1)−f(x)$
So for this problem, a basis for $P_3$ is $\{1, x, x^2, x^3\}$, but looking at this problem I'm not sure that is even necessary...? I really don't know where to start. 
 A: $$f(x)=ax^3+bx^2+cx+d\in\ker \phi\Longleftrightarrow $$
$$a(x+1)^3+b(x+1)^2+c(x+1)+d-(ax^3+bx^2+cx+d)=0\Longleftrightarrow$$
$$3ax^2+(3a+2b)x+(a+b+c)=0\Longleftrightarrow a=0\,,\,b=c\,,\,b,d\in\Bbb Z/2\Bbb Z$$
since $\,3a+2b=0\,\,\wedge\,\,a=0\Longrightarrow 2b=0\,$ , but $\,2=0\,$ here, so $\,b\,$ can be whatever.
Can you take it from here?
A: Let us compute the matrix of the operator $T(f)(x)=f(x+1)-f(x)$ with respect to the canonical basis.
We find:
$$
\left(\matrix{0&1&1&1\\0&0&2&3\\0&0&0&3\\0&0&0&0}\right)=\left(\matrix{0&1&1&1\\0&0&0&1\\0&0&0&1\\0&0&0&0}\right).
$$
We first see, by solving the appropriate system, that
$$
\mbox{Ker}\;T=\{a+bx+bx^2\;;\;a,b\in \mathbb{Z}/2\mathbb{Z}\}
$$
Basis: $\{1,x+x^2\}$.
Then we easily see that
$$
\mbox{Im}\;T=\{a+bx+bx^2\;;\;a,b\in \mathbb{Z}/2\mathbb{Z}\}.
$$
Basis: $\{1,x+x^2\}$.
So the kernel and the range are equal.
Note: we could have observed from the beginning that $T^2=0$. So the range is contained in the kernel. So it would have been sufficient to determine the kernel, whose dimension is $2$, to conclude that the two subspaces are equal via the rank-nullity theorem.
