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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 $T$.

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  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Nov 22 '16 at 5:03
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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. You might prefer to use the fact that if the minimal polynomial is smaller than the characteristic, then there is an eigenspace of dimension at least $2$.

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  • $\begingroup$ Some alternative arguments regarding that last implication are presented here. $\endgroup$ – Omnomnomnom Nov 22 '16 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 Nov 22 '16 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 Nov 22 '16 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). $\endgroup$ – Omnomnomnom Nov 22 '16 at 14:14

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