Let $\operatorname{ad}_x:L\rightarrow GL(L)$ be the adjoint representation. In Humphreys "Introduction the Lie-Algebras and representation theory" one can find this example where $x,y$ and $h$ are defined here. (The first picture can be find on page 6 and the second one on page 22).
I would like to understand how one can compute $\operatorname{ad}_x$. Hopefully someone can help me :)
For a Lie algebra $\mathfrak{g}$ The function $ad_x: \mathfrak{g} \to \mathfrak{g}$ is given by $ad_x(y) = [x,y]$. Often, with matrices, $[x,y] = xy - yx$. So if $\mathfrak{g} = \mathfrak{sl}_2$ and $$ x = \pmatrix{0 & 1 \\ 0 & 0} $$ and $$ y = \pmatrix{a & b \\ c & -a} \in \mathfrak{g} $$ you will have $$ ad_x(y) = \pmatrix{0 & 1 \\ 0 & 0}\pmatrix{a & b \\ c & -a} - \pmatrix{a & b \\ c & -a}\pmatrix{0 & 1 \\ 0 & 0} = \dots $$
The Lie brackets of $\mathfrak{sl}(2)$ are given by $[x,y]=z$, $[z,x]=2x$ and $[z,y]=-2y$. The adjoint map $ad(x)$ is the linear map $T$ with $T(x)=[x,x]=0$, $T(y)=[x,y]=z$ and $T(z)=[x,z]=-2x$. As usual, as a matrix, the image vectors are the columns of the matrix with respect to the basis $x=(1,0,0)$, $y=(0,1,0)$ and $z=(0,0,1)$. So the columns of $T$ are $(0,0,0)^T$, $(0,0,1)^T$ and $(-2,0,0)$, i.e., $$ T=ad(x)=\begin{pmatrix} 0 & 0 & -2 \cr 0 &0 & 0 \cr 0 & 1 & 0\end{pmatrix}. $$ Similarly we obtain $ad(y)$ and $ad(z)$. For a similar question see here.
