2

I want to prove the following statemaent.

Let $V$ be a $n$-demensional vector space on field $K$ and $f:V\rightarrow V$ be a linear operator. There exists $v\in V$ such that {$v,f(v),f^2(v),\ldots,f^{n-1}(v)$} is a basis of $V$ if the minimal polynomial of $f$ is equal to the characteristic polynomial of $f$.

I've checked this page, but I couldn't understand why $v_i, Tv_i, T^2v_i, \ldots, T^{\mu_{j}-1}v_i$ are linearly independent (in the answer by Yiorgos S. Smyrlis).

Any help is appreciated. Thanks.

Note that $K$ is not neseccarily a algebraically closed field.

kufs
  • 21
  • You might find my post here to be useful. The "cyclic vector" $v$ that I construct is the vector for which ${v,f(v),\dots,f^{n-1}(v)}$ is a basis. – Ben Grossmann Jul 20 '20 at 13:57
  • $K$ is not necessarily algebraically closed field, so $f$ do not necessarily have Jordan form. – kufs Jul 21 '20 at 08:23
  • That's a fair point. That said, the page you link has the same issue, then. – Ben Grossmann Jul 21 '20 at 08:25
  • I believe this post provides a suitable argument. – Ben Grossmann Jul 21 '20 at 08:30
  • Alternatively, if you can appeal to the existence of Frobenius canonical form or the structure theorem for finitely generated module over a PID (cf. Hoffmann and Kunze or Dummit and Foote respectively), this statement is an immediate consequence. – Ben Grossmann Jul 21 '20 at 08:34
  • I tried this, but I can't : If $P(f)=a_0id_V+a_1f+\ldots+a_{n-1}f^{n-1}(a_0, a_1, \ldots, a_{n-1}\in K)$ is a zero mapping when $a_0w+a_1f(w)+ \ldots+a_{n-1}f^{n-1}(w)=(a_0id_V+a_1f+\ldots+a_{n-1}f^{n-1})(w)=0 ()$, then $a_0=a_1=\ldots=a_{n-1}=0$. It is because if at least one of $a_0, a_1, \ldots, a_{n-1}$ is not $0$, $P(t)=a_0 +a_1 t+ \ldots+a_{n-1}t^{n-1}\neq0$, which means $0\neq P(t)\in { F(t)\in K[t]|F(f)=0}$　but $degP(t)<deg\varphi(t)(\varphi(t):$ the minimal polynomial of $f)$. This is contradiction, and $()$ don't have any nontrivial solutions. – kufs Jul 21 '20 at 09:05
  • I don't understand what you're trying to accomplish in this comment – Ben Grossmann Jul 21 '20 at 09:07
  • If $a_0w+a_1f(w)+\ldots+a_{n-1}f^{n-1}(w)=0$ don't have any nontrivial solutions, $w,f(w),\ldots,f^{n-1}(w)$ are linearly independent. – kufs Jul 21 '20 at 09:10
  • There is a difference between saying "$a_0w+a_1f(w)+\ldots+a_{n-1}f^{n-1}(w)=0$ has no trivial solutions" and "$a_0 \text{id}+a_1f+\ldots+a_{n-1}f^{n-1}=0$ has no trivial solutions." The second statement is, by itself, not enough to conclude that the first statement holds for some $w$. – Ben Grossmann Jul 21 '20 at 09:16
  • Also, I just gave you a link to a post that explains an approach to your problem. Is there something unclear or unsatisfactory about this solution? – Ben Grossmann Jul 21 '20 at 09:18
  • I now understand this page . I'm sorry for being late to answer you. Thanks. – kufs Jul 21 '20 at 09:48
  • Not a problem, glad you understand it. If anything about that page is unclear, I could try to clarify it in an answer below, or you could consider posting a new question. – Ben Grossmann Jul 21 '20 at 09:50
  • Please let me take back waht I said. Why is ${T^k(v_i)|0\leqq k<m_idegP_i}$ a basis of $V_i$ when $P^{m_i-1}[T]v_i\neq0?$ – kufs Jul 21 '20 at 10:22

1 Answers1

1

Regarding your comment: I will remove the subscript $i$ to simplify notation.

At the relevant point in the proof, we are given that $T|_{V}$ has minimal polynomial $P^m$, where $P$ is irreducible, and we must show the existence of a cyclic generator of $V$, i.e., a vector $v\in V$ such that the minimal degree monic polynomial $Q$ such that $Q[T](v)=0$ has $\deg(Q)=\dim(V)$.

Let $d = \dim(V)$. Note in particular that the degree of this polynomial $Q$ is the dimension of the span of $\{v,Tv,\dots,T^{d-1}v\}$.

As the minimal polynomial of $T|_{V}$ is $P^{m}$, this polynomial $Q$ divides $P^{m}$ for any vector$~v\in V$: only powers of $P$ can occur. It therefore suffices to find a vector$~v\in V$ for which $P^{m-1}[T](v)\neq0$: if $P^{m-1}[T](v)\neq0$, then we know that $Q$ divides $P^m$ but does not divide $P^{m-1}$, which means that we must have $Q = P^m$. However, that means that $\deg(Q) = \deg(P^m) = d$, which means that the set $\{v,Tv,\dots,T^{d-1}v\}$ spans $V$.

Ben Grossmann
  • 225,327