Let \(V\) be an \(n\)-dimensional vector space over the field \(F\), and let \(\mathfrak{B} = \{\alpha_1, \ldots, \alpha_n\}\) be an ordered basis for \(V\).
According to Theorem 1, there is a unique linear operator \(T\) on \(V\) such that \[ T\alpha_j = \alpha_{j+1}, \quad j = 1, \ldots, n-1, \quad T\alpha_n = 0. \] What is the matrix \(A\) of \(T\) in the ordered basis \(\mathfrak{B}\)?
Prove that \(T^n = 0\) but \(T^{n-1} \neq 0\).
Let \(S\) be any linear operator on \(V\) such that \(S^n = 0\) but \(S^{n-1} \neq 0\). Prove that there is an ordered basis \(\mathfrak{B}'\) for \(V\) such that the matrix of \(S\) in the ordered basis \(\mathfrak{B}'\) is the matrix \(A\) of part (a).
Prove that if \(M\) and \(N\) are \(n \times n\) matrices over \(F\) such that \(M^n = N^n = 0\) but \(M^{n-1} \neq 0 \neq N^{n-1}\), then \(M\) and \(N\) are similar.
Solution
For part (a), the \(j\)th column of the matrix is the coordinate vector of \(T\alpha_j\). Hence \[ A=\begin{bmatrix} 0&0&\cdots&0&0\\ 1&0&\cdots&0&0\\ 0&1&\ddots&0&0\\ \vdots&\ddots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0 \end{bmatrix}. \] It has ones immediately below the main diagonal.
For part (b), \(T^k\alpha_j=\alpha_{j+k}\) when \(j+k\le n\) and is zero otherwise. Hence \(T^n=0\), while \[ T^{n-1}\alpha_1=\alpha_n\ne0. \]
For part (c), choose \(v\in V\) with \(S^{n-1}v\ne0\). We claim \[ \mathfrak B'=\{v,Sv,\ldots,S^{n-1}v\} \] is a basis. If \[ c_1v+c_2Sv+\cdots+c_nS^{n-1}v=0, \] apply \(S^{n-1}\) to obtain \(c_1S^{n-1}v=0\), so \(c_1=0\). Applying successively \(S^{n-2},S^{n-3},\ldots\) shows every \(c_j=0\). Thus these \(n\) vectors are independent and form a basis. In this basis \(S\) sends each basis vector to the next and the last to zero, so its matrix is \(A\).
For part (d), the operators defined by \(M\) and \(N\) both admit a basis in which their matrices are \(A\). Therefore \(M\) and \(N\) are each similar to \(A\), hence similar to each other.