= Solution
Every $2\times2$ <matrix> has the unique decomposition
$$
A=\frac{\operatorname{tr}A}{2}I+\left(A-\frac{\operatorname{tr}A}{2}I\right).
$$
Both summands define invariant subspaces for the <Adjoint representation of a Lie algebra>. Scalar matrices commute with everything, so $\mathbb CI$ is $L_0$. The traceless matrices form $\mathfrak{sl}_2$ with <basis> $e,h,f$ of weights $2,0,-2$. Their brackets satisfy $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$; the raising and lowering operations connect all three weight lines. This is the <simple module> $L_2$. Therefore
$$
\boxed{\mathfrak{gl}_2(\mathbb C)\cong L_0\oplus L_2.}
$$
As a check, its zero-<weight space> has <dimension> two and its one-<weight space> is zero, agreeing with part (ii).
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