Let $V$ be an $n$-dimensional vector space over the field $F$, and let $N$ be a nilpotent linear operator on $V$. Suppose $N^{n-1}\neq 0$, and let $\alpha$ be any vector in $V$ such that $N^{n-1}(\alpha)\neq 0$. Prove that $\alpha$ is a cyclic vector for $N$. What exactly is the matrix of $N$ in the ordered basis $\{\alpha, N(\alpha ),…, N^{n-1}(\alpha)\}$?
Approach (1): We need to show $V=Z(\alpha; N)=\text{span}(\{N^k(\alpha)\mid k\geq 0\})$. Let $B=\{\alpha,N(\alpha),…,N^{n-1}(\alpha)\}$. Since $N^n(\alpha)=0$, we have $N^{k}(\alpha)=0$, $\forall k\geq n$. So $\text{span}(\{N^k(\alpha)\mid k\geq 0\})=\text{span}(\{N^k(\alpha)\mid 0\leq k\leq n-1\})$. Thus $Z(\alpha;N)=\text{span}(B)$. Suppose $c_0\cdot \alpha +c_1\cdot N(\alpha)+…+c_{n-1}\cdot N^{n-1}(\alpha)=0$, for some $c_i\in F$. Then $N^{n-1}(c_0\cdot \alpha +c_1\cdot N(\alpha)+…+c_{n-1}\cdot N^{n-1}(\alpha))=c_0\cdot N^{n-1}(\alpha)=0$. Since $N^{n-1}(\alpha)\neq 0$, we have $c_0=0$. Similarly $c_i=0$, $\forall i\in J_{n-1}$. So $B$ is independent. Thus $B$ is basis of $Z(\alpha;N)$. Since $Z(\alpha ;N)$ is subspace of $V$ and $\dim Z(\alpha;N)=n=\dim(V)$, we have $Z(\alpha;N)=V$. It’s easy to check, $[N]_B=\begin{bmatrix} & & & \\ 1 & & & \\ &\ddots & & \\ &&1& \\\end{bmatrix}$.
Approach (2): Let $p_\alpha$ be monic generator of ideal $S_N(\alpha ;\{0\})$, $p_\alpha$ is also called $N$-annihilator of $\alpha$. By linear independence of $B=\{\alpha,N(\alpha),…,N^{n-1}(\alpha)\})$, we have $\deg (p_\alpha)\geq n$. Since $N^n(\alpha)=0$ and uniqueness of monic generator, $p_\alpha=x^n$. By theorem 1 section 7.1, $\deg (p_\alpha)=n=\dim Z(\alpha ;N)$. Thus $Z(\alpha ;N)=V$. Is my proof correct?
I think, we don’t need nilpotent property on $N$, only $N^{n-1}(\alpha)\neq 0$ and $N^n(\alpha)=0$.
