I'm trying to understand the proof of Cartan's Criterion for Solvability given here, and have two questions: On page 15, about half way down, we assert the following:
If $\mathfrak{g}=\mathfrak{g}_0 \oplus \bigoplus_{\alpha \in \Phi}\mathfrak{g}_{\alpha}$ is the Cartan Decomoposition, where $\mathfrak{g}_0=\mathfrak{h}$ is the Cartan subalgebra, and $(V,\rho)$ is a finite dimensional representation of $\mathfrak{g}$ with a decomposition into generalised eigenspaces $V= \bigoplus_{\mu \in \Psi}V_{\mu}$ as a $\mathfrak{h}-$representation then:
If $V=V_0$, then clearly this means that any $x \in \mathfrak{h}$ acts nilpotently as an endomorphism of $V$ (i.e. $\rho(x)$ is nilpotent. However, it is further asserted that this in fact implies that $x$ is ad-niplotent. I can't see where this is coming from: if $y \in \mathfrak{g}$, then $ad(x)^n(y)=$ $[x,[x,[...[x,y]]..]$ , and is there something to with viewing $x$ and $y$ here as endomorphisms on $V$ via the representation?
Why in the second step is it not okay to stop at the statement that
$$ t_V(x,x)=tr(x^2)=\sum_{\mu \in \Psi}\dim V_{\mu}.\mu(x)^2$$ since we know each of the terms on the RHS is $\ge 0$, and that at least one of them (where $\mu = \lambda$) is non-zero because $\lambda \in \Psi$ so $V_{\lambda}\ne 0$ and $\lambda(x) \ne 0$.