Compute the adjoint representation of $\SU(2)$ explicitely.

For any $U=(u_{jk})\in\SU(2)$ we get $$ \Ad\left(\begin{array}{cc} u_{11}&u_{12}\\ u_{21}&u_{22} \end{array}\right) \left(\begin{array}{cc} a&b\\ c&-a \end{array}\right) =\left(\begin{array}{cc} u_{12}(c\bar u_{22}+a\bar u_{21})+u_{11}(a\bar u_{22}-b\bar u_{21}) &u_{12}(-c\bar u_{12}-a\bar u_{11})+u_{11}(b\bar u_{11}-a\bar u_{12})\\ u_{22}(c\bar u_{22}+a\bar u_{21})+u_{21}(a\bar u_{22}-b\bar u_{21}) &u_{22}(-c\bar u_{12}-a\bar u_{11})+u_{21}(b\bar u_{11}-a\bar u_{12}) \end{array}\right)~. $$ In particular for $H:a=1,b=c=0$, $X:b=1,a=c=0$ and $Y:c=1,a=b=0$: $$ \Ad(U)H=\left(\begin{array}{cc} u_{12}\bar u_{21}+u_{11}\bar u_{22} &-u_{12}\bar u_{11}-u_{11}\bar u_{12}\\ u_{22}\bar u_{21}+u_{21}\bar u_{22} &-u_{22}\bar u_{11}-u_{21}\bar u_{12} \end{array}\right)~. $$ and $$ \Ad(U)X=\left(\begin{array}{cc} -u_{11}\bar u_{21} &u_{11}\bar u_{11}\\ -u_{21}\bar u_{21} &u_{21}\bar u_{11} \end{array}\right)\quad \Ad(U)Y=\left(\begin{array}{cc} u_{12}\bar u_{22} &-u_{12}\bar u_{12}\\ u_{22}\bar u_{22} &-u_{22}\bar u_{12} \end{array}\right)~. $$