Question 4.E.8.11: N is nilpotent of index k when N^k = 0 but N^k−1 ≠ 0. If N i......
(a) N is nilpotent of index k when N^k = 0 but N^{k−1} ≠ 0. If N is a nilpotent operator of index n on ℜ^n, and if N^{n−1}(y) ≠ 0, show \mathcal{B} = \{y, N(y), N^2(y), . . . , N^{n−1}(y)\} is a basis for ℜ^n, and then demonstrate that
[N]_{\mathcal{B}} = J = \begin{pmatrix}0 &0& · · · &0 &0\\1 &0& · · ·& 0& 0\\ 0 &1 &· · ·& 0 &0\\\vdots&\vdots&\ddots&\vdots&\vdots\\0 &0& · · ·& 1& 0\end{pmatrix}.
(b) If A and B are any two n × n nilpotent matrices of index n, explain why A \simeq B.
(c) Explain why all n × n nilpotent matrices of index n must have a zero trace and be of rank n − 1.
(a) Because \mathcal{B} contains n vectors, you need only show that \mathcal{B} is linearly independent To do this, suppose \sum_{i=0}^{n−1} α_iN^i(y) = 0 and apply N^{n−1} to both sides to get α_0N^{n−1}(y) = 0 \Longrightarrow α_0 = 0. Now \sum_{i=1}^{n−1} α_iN^i(y) = 0. Apply N^{n−2} to both sides of this to conclude that α_1 = 0. Continue this process until you have α_0 = α_1 = · · · = α_{n−1} = 0.
(b) Any n × n nilpotent matrix of index n can be viewed as a nilpotent operator of index n on ℜ^n. Furthermore, A = [A]_{\mathcal{S}} and B = [B]_{\mathcal{S}}, where \mathcal{S} is the standard basis. According to part (a), there are bases \mathcal{B} and \mathcal{B}^′ such that [A]_{\mathcal{B}} = J and [B]_{\mathcal{B}^′} = J. Since [A]_{\mathcal{S}} \simeq [A]_{\mathcal{B}}, it follows that A \simeq J. Similarly B \simeq J, and hence A \simeq B by Exercise 4.8.2.
(c) Trace and rank are similarity invariants, and part (a) implies that every n × n nilpotent matrix of index n is similar to J, and trace (J) = 0 and rank (J) = n − 1.