I'm reading through Mukai's excellent book "Introduction to Invariants and Moduli", and am stuck on a proof in Chapter 4. He's proving that $G = SL_n$ over a field $k$ of characteristic $0$ is linearly reductive, i.e. for every epimorphism $V \rightarrow W$ of representations of $G$, the induced map on invariants $V^G \rightarrow W^G$ is also surjective. Let $\rho$ be a representation of $SL_n$ and let $\tilde{\rho}$ be the induced representation on the Lie algebra and the distribution algebra at the identity. Let $\Omega$ be the Casimir element/operator. Let $T$ be the torus of diagonal matrices in $SL_n$ and let $\frak{h}$ be its Lie algebra.

Mukai reduces the proof of linear reductivity to the following assertion: if $\mathrm{tr}( \tilde{\rho}(\Omega)) = 0$ we must also have $\mathrm{tr}(\tilde{\rho}(h)) = 0$ for all $h \in \frak{h}$. He then says: we will do this just for $SL_2$; the general case is similar. For $SL_2$ we have $\frak{h}$ is one-dimensional spanned by multiples of the root $h = \epsilon_1 - \epsilon_2$, and by explicit calculation $\mathrm{tr}( \tilde{\rho}(\Omega)) = \mathrm{tr}(\tilde{\rho}(h)^2)$. So the assertion is immediate. But I don't quite see how the "similar" proof for $SL_n$ works. It would be great if someone could explain this to me!