# Prove that $T$ has a cyclic vector iff its minimal and characteristic polynomials are the same

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 $T$.

• 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. Commented Nov 22, 2016 at 5:03

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.

• Some alternative arguments regarding that last implication are presented here. Commented Nov 22, 2016 at 5:26
• Not very clear where to start, but thank you so much. I will try to expand the idea into full understanding of the problem. Commented Nov 22, 2016 at 5:46
• 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. Commented Nov 22, 2016 at 14:10
• 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). Commented Nov 22, 2016 at 14:14

Let dim(V) = n. The minimal polynomial is equal to the characteristic polynomial if and only if $$$$n = \min \{ deg(f) | f \in F[x] \text{ s.t.} f(T)= 0 \},$$$$ if and only if $$$$\forall f\in F[x], \text{ s.t. } deg(f) =n-1 \text{ and } f(T)=0, \,\,\Rightarrow f = 0$$$$ if and only if $$$$\sum_{i=0}^{n-1} a_{i}T^{i} = 0, \Rightarrow \,\, a_{i}=0, i=0, 1, \dots, n-1.$$$$ The latter is equivalent to the existence of a $$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.

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