Suppose that there is a basis of $V$ such that, with respect to this basis, every $x \in L$ is represented by a strictly upper triangular matrix….

Question

This is exercise 5.4 in Erdmann and Wildons Introduction to Lie Algebras.

Let $L$ be a Lie Subalgebra of $\mathfrak{gl}(V)$. Suppose that there is a basis of $V$ such that, with respect to this basis, every $x \in L$ is represented by a strictly upper triangular matrix. Show that $L$ is isomorphic to a Lie subalgebra of $n(n,F)$ and hence that $L$ is nilpotent.

$n(n,F)$ is the strictly upper triangular matrices.

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My attempt:

So, let $B$ be the basis in which every element of $L$ is represented by a strictly upper triangular matrix.

Let $\beta: \mathfrak{gl}(V) \rightarrow M_n(\mathbb{C})$ be the canonical isomorphism that we get whenever we choose a basis for a vector space, in this case $B$. Then the image of $L$ will be a Lie sub algebra of $M_n(\mathbb{C})$, i.e. $\beta(L) \leq M_n(\mathbb{C})$. Furthermore, since all $x \in L$ are represented as a strictly upper triangular matrix with basis $B$, we have that $\beta(L) \leq n(n,F)$

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Is this right? I feel like this exercise was painfully obvious, and i'm wondering if there is some hidden complexity that I was not aware of.

Answer 1

Your solution seems correct. Another way to see that $L$ is nilpotent is to use Engel's theorem, which I assume the book is setting up with this exercise.