 Solution

Use the same nonsemisimple matrix from part (i) and the one-dimensional Lie subalgebraIt is a Lie subalgebra because $[aX,bX]=0$ for all scalars $a,b$. Its Jordan–Chevalley decomposition has $X_s=I_2$ and $X_n=X-I_2\ne 0$. If $I_2=cX$, comparing off-diagonal entries forces $c=0$, which contradicts the diagonal entries. If $X_n=cX$, taking traces gives $0=2c$, again forcing $c=0$ although $X_n\ne 0$. ThusThe semisimplicity assumption on the Lie algebra in part (ii) is therefore essential.