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](http://projecteuclid.org/euclid.bams/1183529092)) 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!




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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 $b\ge n$.



Lie algebra admitting some hyperbolic automorphism is nilpotent
First, if $x$ and $y$ are generalised eigenvectors of $\phi$ with generalised eigenvalues $\alpha$ and $\beta$, that is $(\phi-\alpha\mathrm{Id}\_\mathfrak{g})^N(x)=(\phi-\beta\mathrm{Id}\_\mathfrak{g})^M(y)=0$, then we observe that $\mathrm{ad}(x)^K(y)=0$ for $K>>0$, just noticing that $\mathrm{ad}(x)^K(y)$ is a generalised eigenvector with generalised eigenvalue $\alpha^K\beta$, so since $\alpha$ is not a root of unity, not all of these have nonzero generalised eigenvectors. Thus, if $x$ is a generalised eigenvector of $\phi$, then $\mathrm{ad}(x)$ is nilpotent.


Here I got stuck and ended up checking Bourbaki; they say that one has to use the following result (which I simplify substantially compared to the general statement of Ex.11 in the same volume): "If $L$ is a subset [not necessarily a subspace!] of the matrix algebra closed under commutators and we know that $L$ consists of nilpotent matrices, then the associative subalgebra of the matrix algebra generated by $L$ is nilpotent." (This seems fairly easy, I am just a bit lazy to type the proof.) This result should, of course, be applied to the subset of $\mathrm{End}(\mathfrak{g})$ consisting of all linear operators $\mathrm{ad}(x)$, where $x$ is a generalised eigenvector of $\phi$.


