1

Suppose we have a linearly independent linear operators $\mathcal{E}, \mathcal{A},\mathcal{A}^2,\ldots, \mathcal{A}^{n-1}$ which acts on finite dimension vector space $V$, $n = \operatorname{dim}V$. Show that there exists a vector $v \in V$ such that $ V = \left<v, \mathcal{A}v, \mathcal{A}^2v,\ldots,\mathcal{A}^{n-1}v \right>$

I only realized that $\mathcal{A} \left(\left<v, \mathcal{A}v, \mathcal{A}^2v,\ldots,\mathcal{A}^{n-1}v \right> \right) = \left<v, \mathcal{A}v, \mathcal{A}^2v,\ldots,\mathcal{A}^{n-1}v \right>$ for all $v \in V $, this follows from the fact that $\mathcal{E}, \mathcal{A},\mathcal{A}^2,\ldots, \mathcal{A}^{n-1}$ lin. independent consequently $\mathcal{A}^n = \sum_{i=0}^n\alpha_i\mathcal{A}^i $. Can it be helpful...

Addition.

Let's consider for any vector $ v\in V $ minimal polynomial of this vector (relative to a fixed operator $ \mathcal{A}$) it is a polynomial $ f_v $ of minimal degree with the leading coefficient equal to $1$ and $ f_v (\mathcal{A})(v) = 0$. It is easy to show that $ f_v \mid \mu_{\mathcal{A}} ,\ \forall v \in V$, and since the polynomial has a finite number of divisors, there exists polynomials $ f_1, \ldots, f_k$ with the property $ \forall v \in V \ \exists i \colon f_v = f_i$. Now let's look at the subspaces in $ V $, $ V_i = \{a \in V \mid f_i(\mathcal{A})(a) = 0\} $, we have $ V = \cup_{i=1}^{k}V_i $. Now all that's left is to show $ V = V_i $ or $\operatorname{deg}f_i = n $ for some $ i $. I'm thinking about it...

victor
  • 39
  • 1
    Pick $v$ from the complement of $\operatorname{ker}(A)$? – Captuna Mar 15 '25 at 09:49
  • @Captuna I had the same thoughts, just to take $ v \notin \operatorname{Ker}\mathcal{A}^{n-1}$. It follows that all vectors are non-zero $\mathcal{A}v, \mathcal{A}^2v,\ldots,\mathcal{A}^{n-1}v$. But are they linearly independent... – victor Mar 15 '25 at 16:03
  • 1
    @victor See this post for one possible approach. It would be unusual that you would be expected to show that this holds directly; perhaps you're meant to use the existence of rational canonical form. – Ben Grossmann Mar 15 '25 at 18:49
  • 1
    @Captuna That doesn't quite work. For example, consider the "Jordan block" $$ A = \pmatrix{0 & 1\&\ddots & \ddots\ && 0 & 1\&&&0} $$ Saying that $v = (v_1,\dots,v_n)$ is in the "complement of $\ker A$" amounts to saying that one of the elements $v_2,\dots,v_n$ are non-zero. However, the vectors $v, Av, \dots, A^{n-1}v$ will only be independent (more specifically, will only be non-zero) if $v_n \neq 0$.

    Interestingly, this is a situation where "most" vectors will be suitable.

     – Ben Grossmann Mar 15 '25 at 18:53
  • 1
    @victor There is also an approach to proving this result using Jordan canonical form, if that would be preferable. – Ben Grossmann Mar 15 '25 at 18:54
  • 1
    @victor See my post here for an approach using Jordan form. – Ben Grossmann Mar 15 '25 at 19:01
  • @BenGrossmann Thank you for answering my question. This exercise is given in the book before studying the Jordan form, so the approaches you sent are still difficult for me, but I will definitely study it. I have another approach, I supplemented my question above, I hope this attempt will be fruitful... – victor Mar 16 '25 at 02:33
  • @victor Which textbook is this from? By the way, the first approach that I linked makes no use of any canonical forms – Ben Grossmann Mar 16 '25 at 03:35
  • @BenGrossmann This is from the russian textbook by A. I. Kostrikin "Introduction to Algebra. Part 2: linear algebra", chapter 2, paragraph 3, exercise 4. I have not found this book in English. – victor Mar 16 '25 at 05:27

0 Answers0