8
$\begingroup$

Let $k$ be an algebraically closed field and $V$ be a finite-dimensional $k$-vector space of dimension $n$.

Let $T:V \rightarrow V$ be a $k$-linear endomorphism of $V$. A vector $v \in V$ is called a cyclic vector for $T$ if the set of vectors $\{T^nv: n \in \mathbb{Z}, n \geqslant 0\}$ span $V$.

1 Show that if $v \in V$ is a cyclic vector, then $\{v, Tv,\cdots, T^{n-1}v\}$ form a basis for $V$.

2 If $T$ admits a cyclic vector, and $A:V\rightarrow V$ is a linear map commuting with $T$, show that there exists a polynomial $P(x) \in k[x]$ such that $A=P(T)$.

3 Show that a cyclic vector for $T$ exists if and only if the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$.

$\endgroup$
1
  • 1
    $\begingroup$ What are your thoughts on the question? Have you tried anything so far? On this site, we expect askers to provide a little bit of context to their problem. $\endgroup$ Commented Nov 22, 2016 at 5:03

2 Answers 2

8
$\begingroup$

Hints:

  1. By the Cayley-Hamilton theorem, $T^kv$ is in the span of $\{v,Tv,\dots,T^{n-1}v\}$ for any $k \geq n$
  2. By 1, there exists a polynomial $T$ of degree $n-1$ such that $$ Av = p(T)v $$ Then, note that $AT^kv = T^k Av = T^kp(T)v = p(T)T^k v$ for $k = 0,1,\dots,n-1$.

  3. If the minimal polynomial is of lower degree, then $T^{n-1}$ is a linear combination of $\{I,T,\dots,T^{n-2}\}$, so there can be no cyclic vector. The opposite implication is tricky; it suffices to consider the Jordan form or rational canonical form of $T$, but that may be excessive.

$\endgroup$
4
  • $\begingroup$ Some alternative arguments regarding that last implication are presented here. $\endgroup$ Commented Nov 22, 2016 at 5:26
  • $\begingroup$ Not very clear where to start, but thank you so much. I will try to expand the idea into full understanding of the problem. $\endgroup$
    – adosdeci
    Commented Nov 22, 2016 at 5:46
  • $\begingroup$ For 1, should I prove that they are linearly independent? I think that's the point of the question. So if not, there might exist a degree at most $n-1$ polynomial which $T$ is a root. $\endgroup$
    – adosdeci
    Commented Nov 22, 2016 at 14:10
  • $\begingroup$ My hint tells you that if $v$ is a cyclic vector, then the set $\{v,Tv,\dots,T^{n-1}v\}$ of $n$ vectors must span $V$, an $n$-dimensional space. In an $n$-dimensional space, any spanning set of $n$ vectors is necessarily a basis (i.e. is necessarily linearly independent). $\endgroup$ Commented Nov 22, 2016 at 14:14
-1
$\begingroup$

Let dim(V) = n. The minimal polynomial is equal to the characteristic polynomial if and only if \begin{equation} n = \min \{ deg(f) | f \in F[x] \text{ s.t.} f(T)= 0 \}, \end{equation} if and only if \begin{equation} \forall f\in F[x], \text{ s.t. } deg(f) =n-1 \text{ and } f(T)=0, \,\,\Rightarrow f = 0 \end{equation} if and only if \begin{equation} \sum_{i=0}^{n-1} a_{i}T^{i} = 0, \Rightarrow \,\, a_{i}=0, i=0, 1, \dots, n-1. \end{equation} The latter is equivalent to the existence of a $ v \neq 0 $ s.t. if $ \sum_{i=0}^{n-1} a_{i}T^{i}(v) = 0 $ then $ a_{0}=a_{1}=\dots = a_{n-1} = 0 $. This is because, if for every nonzero $ v $, $ T^{i}(v) = 0 $, for $ i = 1, 2, \dots , n-1 $, then for $ f(x) = x + x^{2} + \dots + x^{n-1} $, $ f(T)= 0 $, and again it contradicts $ m_{T}(x) = \rho_{T}(x) $. We conclude that, $ m_{T}(x) = \rho_{T}(x) $ if and only if there exists $ v \neq 0 $ s.t. $ \{ T^{i}(v) \}_{i=0}^{n-1} $ is a basis for $ V $, i.e., $ V $ is $ T $-cyclic.

$\endgroup$
1
  • 1
    $\begingroup$ You did not negate the statement correctly when proving the claim $\exists v\neq 0,\sum a_iT^i=0\Rightarrow a_i=0$. $\endgroup$
    – Hyacinth
    Commented Sep 25, 2022 at 5:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .