I am trying to work out the adjoint representations of
$$H=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right), X = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right), Y = \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right) .$$
but I am not exactly sure how to do it.
So far I have worked out
(ad$H$)($H$) = $0$
(ad$H$)($X$) = $\left(\begin{array}{cc} 0 & 2 \\ 0 & 0 \end{array} \right)$
(ad$H$)($Y$) = $\left(\begin{array}{cc} 0 & 0 \\ -2 & 0 \end{array} \right)$
Let $$ E_{11}=\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right), E_{12}=\left( \begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right), E_{21}=\left( \begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right),\quad \text{and}\quad E_{22}=\left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right).$$ This is a basis of $\mathbb M_{2}$. For each pair $(i,j)$ express $adH(E_{ij})$ as a linear combination of matrices from the basis. For instance $$ adH(E_{12})=2 E_{12}.$$ More generally, $$ adH(E_{ij})=\alpha_{1}^{(ij)}E_{11}+\alpha_{2}^{(ij)}E_{12}+\alpha_{3}^{(ij)}E_{21}+\alpha_{4}^{(ij)}E_{22}. $$ Then $ad H$ has a matrix representation (with respect to the above basis): $$ adH=\left( \begin{array}{cccc} \alpha_{1}^{(11)} & \alpha_{1}^{(12)} & \alpha_{1}^{(21)} & \alpha_{1}^{(22)}\\ \alpha_{2}^{(11)} & \alpha_{2}^{(12)} & \alpha_{2}^{(21)} & \alpha_{2}^{(22)}\\ \alpha_{3}^{(11)} & \alpha_{3}^{(12)} & \alpha_{3}^{(21)} & \alpha_{3}^{(22)}\\ \alpha_{4}^{(11)} & \alpha_{4}^{(12)} & \alpha_{4}^{(21)} & \alpha_{4}^{(22)} \end{array} \right). $$