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 \(\{\alpha, N\alpha, \ldots, N^{n-1}\alpha\}\)?
Solution
Since \(N\) is nilpotent on an \(n\)-dimensional space, its characteristic polynomial is \(x^n\), and Cayley–Hamilton gives \[ N^n=0 \] [@HK71].
1. The vectors \(\alpha,N\alpha,\ldots,N^{n-1}\alpha\) are linearly independent.
Suppose \[ \sum_{j=0}^{n-1}c_jN^j\alpha=0. \] If some coefficient is nonzero, let \(r\) be the least index with \(c_r\ne0\). Apply \(N^{n-1-r}\) to the relation. The \(r\)th term becomes \[ c_rN^{n-1}\alpha, \] while every term with \(j>r\) contains the power \[ N^{n-1-r+j}, \] whose exponent is at least \(n\), and hence vanishes. Thus \[ c_rN^{n-1}\alpha=0. \] The hypothesis \(N^{n-1}\alpha\ne0\) forces \(c_r=0\), a contradiction. Therefore all coefficients vanish, so the \(n\) displayed vectors are linearly independent. Since \(\dim V=n\), they form a basis. Hence \(\alpha\) is cyclic.
2. In the stated ordered basis, \(N\) is the nilpotent shift matrix.
Let \[ \mathfrak B=(\alpha,N\alpha,\ldots,N^{n-1}\alpha). \] For \(0\le j<n-1\), \[ N(N^j\alpha)=N^{j+1}\alpha, \] the next basis vector, while \[ N(N^{n-1}\alpha)=N^n\alpha=0. \] Therefore \[ \boxed{ [N]_{\mathfrak B}= \begin{pmatrix} 0&0&0&\cdots&0\\ 1&0&0&\cdots&0\\ 0&1&0&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0&\cdots&0&1&0 \end{pmatrix}.} \] The ones lie on the subdiagonal because matrix columns are coordinate vectors of the images of the ordered basis vectors.