Nilpotent linear map $x$ induces another nilpotent linear map $ \bar{x} : V/ U \rightarrow V/ U $ that has a basis that is strictly upper triangular.

Question

I'm having trouble proving an exercise (Exercise 6.1 in Erdmann and Wildon's book Introduction to Lie Algebras). The exercise is used to help prove a version of Engel's Theorem.

It states:

Let $V$ be an $n$-dimensional vector space where $ n \geq 1 $ and let $ x : V \rightarrow V $ be a nilpotent linear map.

(i): Show that there is a non-zero vector $v \in V $ such that $ xv = 0 $. (this part I've done).

(ii): Let $U = \text{Span} \{ v \} . $ Show that $ x $ induces a nilpotent linear transformation $ \bar{x} : V/U \rightarrow V/U$. (this part again I've showed and proved it's well defined).

This is the part of the question I'm struggling with:

By induction we know that there is a basis $ \{ v_1 + U , ..., v_{n-1} +U \} $ of $ V/ U $ in which $ \bar{x} $ has a strictly upper triangular matrix. Prove that $ \{ v_1,...,v_{n-1} \} $ is a basis of $ V$ and that matrix of $x$ is strictly upper triangular w.r.t this basis.

I'm having an issue with the induction and the proof from there onwards.

I don't really know where to start, for my $ \bar{x} $ I have $ \bar{x} = x + U $. Any help in the right direction would be appreciated, don't know if I'm missing something obvious.

Answer 1

By induction we know that there is a basis $ \{ v_1 + U , ..., v_{n-1} +U \} $ of $ V/ U $ in which $ \bar{x} $ has a strictly upper triangular matrix.

Let's look at the very first step. Suppose we extend to $v$ to a basis (an arbitrary basis) $\mathcal B = \{v,v_1,\dots,v_{n-1}\}$. We note that the matrix of $x$ with respect to $\mathcal B$ has the form $$ [x]_{\mathcal B} = \left[ \begin{array}{c|ccc} 0 & *\\ \hline 0 & [\bar x]_{\bar{\mathcal B}} \end{array} \right] $$ Where $\bar {\mathcal B} = \{v_1 + U, \cdots, v_{n-1} + U\}$. However, we know that $\bar x$ is itself nilpotent. It follows that we can modify our choice of $v_1,\dots,v_{n-1}$ (so that $v_1 + U \in \ker \bar x$) to get $$ [\bar x]_{\bar{\mathcal B}} = \left[ \begin{array}{c|ccc} 0 & *\\ \hline 0 & [\bar{\bar x}]_{\bar{\bar{\mathcal B}}} \end{array} \right] $$ With respect to this choice of $v_i$, the original matrix has the form $$ [x]_{\mathcal B} = \pmatrix{0&*&*\\0&0&*\\0&0&[\bar{\bar x}]_{\bar{\bar B}}} $$ The pattern continues.

Answer 2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Write $$ V_{i} = \Span{ v_{i}, \dots, v_{n-1}, v}. $$ You know that $\bar x$ maps $V_{i}/U$ to $V_{i+1}/U$ for $i = 1, \dots, n-1$, where $V_{n} = U$. Thus if $v \in V_{i}$, for $i = 1, \dots, n-1$, we have $$ x v + U = \bar x (v + U) \in V_{i+1}/U, $$ so that for $v \in V_{i}$ we have $x v \in V_{i+1}$. And then $x V_{n} = x U = \Set{0}$.