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 $f$ be a linear operator on $\Bbb{K}^4$ whose matrix is $$\begin{bmatrix} c & 0 & 0 & 0 \\ 1 & c & 0 &0 \\ 0 &1 &c & 0 \\0& 0 & 1 & c \end{bmatrix}$$

Let $W=\ker(f-cI_4)$

Find the generators of $S(e_4,W) ,S(e_3,W), S(e_2,W),S(e_1,W)$. Where $S(e_i,W)=\{P(x)\in \Bbb{K}[x] \mid P(f)(e_i)\in W\}$ and $e_i$ are the standard basis vectors.

What I've done: I showed that $W$ is spanned by $e_4$. So for $S(e_4,W)$ , its the set of consant times $e_4$, is it right to say $S(e_4,W)=<c>$?

As for $S(e_3,W)$ : $(f-cI_n)(e_3)=e_4$ hence $S(e_3,W)$ is the set of all polynomials in $\Bbb{K}[x]$ of the form $a_1x+\ldots+a_nx^n$ . I still don't know how to write or find the generator.

As for $S(e_3,W)$ and $S(e_4,W)$ we have $(f-cI_n)(e_2)=e_3$ and $(f-cI_n)(e_1)=e_2$ so upon another application of $(f-cI_n)$ we get $e_4$ . My problem is that I don't know how to write the generators of these sets, especially for the last three sets.

share|cite|improve this question
$f-cI$ has last two columns zero. Its kernel cannot be spanned by $\mathbf{e}_4$ since the kernel is two dimensional. What is true, however, is that $\mathbf{e}_4$ (and $\mathbf{e}_3$) is in $W$. This is enough to conclude that every polynomial is in $S(\mathbf{e}_4,\ W)$ because $P(f)$ commutes with $f-cI$ for all $P\in \Bbb{K}[x]$. Therefore the monic generator of $S(\mathbf{e}_4,\ W)$ (and likewise for $S(\mathbf{e}_3,\ W)$) is $1$. – EuYu Apr 4 '13 at 23:28
@EuYu I am very sorry I got no answers on this so I did not notice the typo . I hope I didn't take much from your time – user10444 Apr 4 '13 at 23:38
up vote 1 down vote accepted

Since $f-cI$ has kernel spanned by $\mathbf{e}_4$ it follows that every polynomial $P\in \mathbb{K}[x]$ is an element of $S$. This is because $P(f)$ commutes with $f-cI$ for all $P$ and therefore $$(f-cI)\left[P(f)\mathbf{e}_4\right] = P(f)\left[(f-cI)\mathbf{e}_4\right] = P(f)\mathbf{0} = \mathbf{0}$$ Therefore the monic generator for $S(\mathbf{e}_4,\ W)$ is $1$ (indeed any non-zero constant can act as a suitable generator as you correctly concluded. We will talk about monic generators for the remainder of the question).

By definition, the generator must divide each element of the ideal. You have actually already done the majority of the work by noticing the shift property of $(f-cI)$.

We have $(x-c)\in S(\mathbf{e}_3,\ W)$ so the generator is either $1$, which is impossible since $\mathbf{e}_3\notin W$, or it is $(x-c)$ which is the case here.

Likewise, we have $(x-c)^2 \in S(\mathbf{e}_2,\ W)$ so the generator is either $1$ (no good since $\mathbf{e}_2\notin W$) or $(x-c)$ (again, no good since $(f-cI)\mathbf{e}_2 = \mathbf{e}_3\notin W$) or $(x-c)^2$, which is the case here.

The case for $S(\mathbf{e}_1,\ W)$ is very similar. Try your hand at that.

share|cite|improve this answer
Thank you very much that cleared a lot up for me, again sorry for the typo. Concerning $S(e_1,W)$ we have $(f-cI)^3e_1=0$ then $(x-c)^3\in S(e_1,W)$ hence the generator is either $1, x-c, (x-c)^2, (x-c)^3$. $1$ does not work since $e_1 \notin W $ neither is it $x-c$ since $(f-cI)e_1=e_2$ neither is it $(x-c)^2$ since $(f-cI)^2e_1=e_3$ hence it is $(x-c)^3$ – user10444 Apr 5 '13 at 0:37
@user10444 That's exactly right! – EuYu Apr 5 '13 at 1:38

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.