if $\phi$ is a linear trasformation on V, its matrix under a group of base vectors is a jordan block, is this V is a cyclic space under $\phi$?

Question

I only know if jordan block is $J_n(0)$ then V will be cyclic space under corresponding linear transformation $\phi$. Will this right for any $J_n(a)\ (a\neq 0)$?

I was really confused while discussing this with my friends.

Answer 1

Hint. $J_n(\lambda)-\lambda I=J_n(0)$.

If a linear operator $A$ acting on a vector space $V$ of dimension $n$ is equivalent to a single Jordan block $J_n(\lambda)$, without loss of generality $V=\mathbb{C}^n$ and $A=J_n(\lambda)$. Then $(A-\lambda I)^ke_n=e_{n-k}$ so $e_n$ is a cyclic generator.