Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
What is the canonical basis? How did you use it to find the matrix? – user4593 Feb 7 '13 at 3:53
That's the basis you gave in your post. For $1$, compute $T(1)$ and write its coefficients in the first column. Then take $x$, compute $T(x)$, write it in the second column. Etc... – 1015 Feb 7 '13 at 3:55

$$f(x)=ax^3+bx^2+cx+d\in\ker \phi\Longleftrightarrow $$


$$3ax^2+(3a+2b)x+(a+b+c)=0\Longleftrightarrow a=0\,,\,b=c\,,\,b,d\in\Bbb Z/2\Bbb Z$$

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?

share|cite|improve this answer
Isn't $2b=0$? In this case, how do you find $b=0$? – 1015 Feb 7 '13 at 3:21
Because I forgot the field was what it is until the end and then I didn't check back. Thanx, I shall edit now. – DonAntonio Feb 7 '13 at 3:22
I think it is actually $a=0$ and $b+c=0$ (or $b-c=0$, equivalently given the field). – 1015 Feb 7 '13 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.