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Problem: Determine the derived series of $\mathfrak{b}_n (\mathbb{C})$, in which $\mathfrak{b}_n (\mathbb{C})$ is the space of all upper triangular matrices.

We knew that the derived series of a Lie algebra $L$ is $L^{(0)}=L, L^{(1)}=[LL],L^{(2)}=[L^{(1)}L^{(1)}], \dots, L^{(i)}=[L^{(i-1)}L^{(i-1)}]$.

How do I finding the derived series of $\mathfrak{b}_n (\mathbb{C})$?

First edited: $L = \mathfrak{b}_n (\mathbb{C})$, so for $n=2$, the basis of $\mathfrak{b}_2 (\mathbb{C})$ is $e_1=\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}, e_2 =\begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix}, e_3=\begin{pmatrix}0 & 0\\ 0 & 1 \end{pmatrix}$

$[e_1 e_2] = e_1, [e_1 e_3]=0, [e_2 e_3]=e_3$

$L^{(1)}=[LL] = \langle e_1,e_3 \rangle$

$L^{(2)}=[L^{(1)} L^{(1)}] = 0$

How to continue the computation?

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marked as duplicate by Dietrich Burde, Torsten Schoeneberg, Robert Soupe, Shailesh, Lord Shark the Unknown Apr 29 at 4:37

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By definition $\mathfrak b_n(\mathbb C)$ is the set of all $n\times n$ complex matrices $(a_{jk})_{1\leqslant j,k\leqslant n}$ such that $a_{jk}=0$ whenever $j>k$. It turns out that, for each $l\in\{1,2,\ldots,n\}$,$$\mathfrak b_n^{(l)}(\mathbb C)=\left\{\text{matrices }(a_{jk})_{1\leqslant j,k\leqslant n}\,\middle|\,j+l>k\implies a_{jk}=0\right\}.$$Prove this by induction on $l$ and you're done.

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