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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.

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2 Answers 2

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.

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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

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

$$a(x+1)^3+b(x+1)^2+c(x+1)+d-(ax^3+bx^2+cx+d)=0\Longleftrightarrow$$

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

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?

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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

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