I am writing a research article in which I need to use the following fact: if $G$ is a subgroup of $GL\_3(\mathbb{R})$ which is irreducible in the sense that no proper nontrivial subspace of $\mathbb{R}^3$ is preserved by all elements of $G$, and if every element of $G$ has all eigenvalues equal in modulus, then there exists $X \in GL\_3(\mathbb{R})$ such that $|\det A|^{-1/3}X^{-1}AX \in O(3)$ for every $A \in G$. Clearly irreducibility is necessary, because otherwise upper-triangular examples exist.


The above result for *semigroups* in arbitrary dimensions follows from Theorem 2 in "Matrix semigroups with constant spectral radius" by V. Yu. Protasov and A. S. Voynov. I however only need the group case, and I find myself wondering whether this can be dealt with more easily. Unfortunately geometry is quite a weak area for me, my main area of expertise being analysis. Can anyone help me by suggesting a relatively quick and painless way to prove the above statement? 


Thanks in advance!



Groups of matrices in which all elements have all eigenvalues equal in modulus
The specific question for $SL(3,\mathbb{R})$ has already been answered by Misha and Uri Bader. I just want to make some additional remarks which say that this holds in greater generality. 


Lemma: Suppose $\Gamma \subset GL(n,\mathbb{C})$ is a subgroup such that all the eigenvalues of all the elements of $G$ have absolute value $1$. Then there exists a closed subgroup $G\subset GL(n,\mathbb{C})$ containing $\Gamma $ which contains a unipotent normal subgroup $U$ such that $G/U$ is compact. 


Clearly, the eigenvalue of any element of such a $G$ will have absolute value $1$, since the eigenvalue of a unipotent element is $1$ and elements of a compact group have eigenvalues of absolute value $1$. Hence the above Lemma gives an if and only if condition for $\Gamma $ to have all eigenvalues of absolute value $1$ . (An alternative way to say this is: $\Gamma $ is contained-not necessarily as a closed subgroup- in a closed amenable subgroup of $GL(n,\mathbb{C})$, which is a unipotent-by- compact extension). 


Proof of the lemma: It is enough to consider the case when $\Gamma$ acts irreducibly on $\mathbb{C}^n$, since the general case follows from this by looking at a $\Gamma $ stable flag in $\mathbb{C}^n$ , where the successive quotients are irreducible. Then the sub-algebra of $M\_n(\mathbb{C})$ generated by the elements $\gamma \in \Gamma$ is all of $M\_n(\mathbb{C})$ by Burnside's theorem. The trace form on $M\_n(\mathbb{C})$ is non-degenerate. Fix a basis of $M\_n(\mathbb {C})$ of the form $g\_i\in \Gamma $ and let $\epsilon \_i$ be the dual basis with respect to the trace form. For every $g\in \Gamma $, we have (see Lemma (2.2) of Tits' article on Free Groups in Journal of Algebra (for a reference see the review


<http://www.ams.org/mathscinet-getitem?mr=286898> ))


$$ g=\sum trace (gg\_i)\epsilon \_i.$$ 
All the elements $g\in \Gamma $ have $bounded$ entries since $trace (gg\_i)$ - being a sum of eigenvalues of $gg\_i$ - is bounded by the dimension $n$. This proves that $\Gamma $ is contained in a compact subgroup of $GL(n, \mathbb{C})$ if it acts irreducibly on $\mathbb{C} ^n$ and all elements have eigenvalues only of norm $1$. 


Now if $\Gamma \subset GL(n, \mathbb{C})$ has the property that the ratio of any two eigenvalues has norm one, then by going to the adjoint representation of $GL(n,\mathbb {C})$ we have: $Ad(\Gamma )$ is contained in an extension of a compact group by a unipotent group. Hence the same holds for $\Gamma $ as well (up to multiplication by scalars). 


