Let $T$ be a linear operator on ${\mathbb R}^3$, which is represented by standard ordered basis, as follows: $$T(e_1) = 2 e_1,$$
$$T(e_2) = 2 e_2,$$
$$T( e_3) = -e_3.$$
I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?
Please suggest !
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$, so they are not the same. Hence $T$ doesn't have cyclic vector by the following theorem:
Theorem: Let $T$ be a linear operator on $n$-dimensional vector space $V$. There exists a cyclic vector for $T$ if and only if minimal polynomial and characteristic polynomial are the same.
Proof: Suppose there exists a cyclic vector $v$ for $T$, that is, we have $v\in V$ such that $\{v, Tv,...,T^{n-1}v\}$ span $V$. Then matrix representation of $T$ will be some companion matrix, whose minimal and characteristic polynomial are the same.
Now conversely, if minimal and characteristic polynomial are the same, then we have a minimal polynomial which is of degree $n$. Take $v\neq 0 $, let minimal polynomial $p(x)= a_0+a_1x+\dots+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider $\{v, Tv, T^2v,...,T^{n-1}v\}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear Algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence the $T$-cyclic subspace generated by $(1,−1,3)$ = linear span of $\{(1,-1,3), (2,-2,-3)\}$.
