Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism.

Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is *hyperbolic* if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.

Then Proposition 3.6 in Smale's paper (here) says that:

- Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

- Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results?

Thanks!

After Vladimir Dotsenko:

$$(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma) v].$$

Applying above to the pair $\hat{u}=\lambda^i\phi^j(\phi-\lambda)^{a}u$ and $\hat{v}=\gamma^k\phi^l(\phi-\gamma)^{b}v$ we have $$(\phi-\lambda\gamma)[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]= [(\phi-\lambda)\hat{u},\phi \hat{v}]+[\lambda \hat{u},(\phi-\gamma) \hat{v}]$$ $$=[\lambda^i\phi^j(\phi-\lambda)^{a+1}u,\gamma^k\phi^{l+1}(\phi-\gamma)^bv] +[\lambda^{i+1}\phi^j(\phi-\lambda)^au, \gamma^k\phi^l(\phi-\gamma)^{b+1}v].$$

Tracing the indices we get $$(i,j,a;k,l,b)\overset{\phi-\lambda\gamma}{\to}(i,j,a+1;k,l+1,b)\cup (i+1,j,a;k,l,b+1),$$ and in particular $(a,b)\overset{\phi-\lambda\gamma}{\to}(a+1;b)\cup (a;b+1)$. Then $$(\phi-\lambda\gamma)^{m+n}[u,v] =\sum_{a+b=m+n}[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=0$$ since either $a\ge m$ or $b\ge n$.