Example: upper triangular matrices (a solvable Lie algebra)

Upper triangular matrices form a Lie algebra whose derived subalgebra is strictly upper triangular, giving an explicit derived series.

Example: upper triangular matrices (a solvable Lie algebra)

Let $B\subset GL(2,\mathbb R)$ (see general linear group ) be the subgroup of invertible upper triangular matrices:

B=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}: a,d\ne 0\right\}.

It is a Lie subgroup, and its Lie algebra is the upper triangular matrices

\mathfrak b=\left\{\begin{pmatrix}x&y\\0&z\end{pmatrix}: x,y,z\in\mathbb R\right\}\subset \mathfrak{gl}(2,\mathbb R).

Commutator calculation

Take

A=\begin{pmatrix}x&y\\0&z\end{pmatrix},\quad A'=\begin{pmatrix}x'&y'\\0&z'\end{pmatrix}.

Then

[A,A']=AA'-A'A = \begin{pmatrix} 0 & (x-z)y'-(x'-z')y\\ 0 & 0 \end{pmatrix},

which is strictly upper triangular.

Therefore the derived subalgebra is

[\mathfrak b,\mathfrak b] = \left\{\begin{pmatrix}0&u\\0&0\end{pmatrix}: u\in\mathbb R\right\} \cong \mathbb R,

matching the strictly upper triangular pattern.

Derived series (explicit)

Since strictly upper triangular $2\times 2$ matrices commute with each other, we get

\mathfrak b^{(1)}=[\mathfrak b,\mathfrak b]=\left\{\begin{pmatrix}0&u\\0&0\end{pmatrix}\right\}, \qquad \mathfrak b^{(2)}=[\mathfrak b^{(1)},\mathfrak b^{(1)}]=0.

Hence $\mathfrak b$ is solvable (see solvable Lie algebra and derived series ).

Context. Upper triangular (Borel) subalgebras are the archetypal solvable subalgebras inside semisimple Lie algebras, and computations like the one above are the linear-algebraic shadow of triangularization phenomena.