Generalization of Jordan's Lemma $A^2=B^2=I$ can be 2-block diagonalized

Question

One of Jordan's lemma states that if two orthogonal matrices $A,B$ are such that $A^2=B^2=I$, then they can be co-diagonalized by block of size 2. (the proof is easy, consider $x$ an eigenvector of $A+B$, $y=Ax$, show that $V=Vect(x,y)$ is stable under $A,B$ and repeat this over the orthogonal of $V$).

Can this be generalized? For instance, to three matrices $A,B,C$ such that $A^3=B^3=C^3=I$ plus some extra hypothesis, which could then be co-diagonalized by block of size 3?

Thanks!

Answer 1

There are a number of possible generalizations, though probably not as general as you might like : here are a couple in the positive direction. The first is very similar to Jordan's result.

1. (H. Blichfeldt): If $V$ is a finite dimensional complex vector space, and $A,B$ are invertible linear transformations on $V$ which each have quadratic minimum polynomial, an are such that $\langle A,B \rangle$ is a finite group, then $V$ is a direct sum of spaces, each of dimension at most two, and each invariant under $\langle A, B \rangle.$

Proof: Let $w$ be an eigenvector of $A-B.$ Then ${\rm span}(w,Aw) = {\rm span}(w,Bw)$ is invariant under both $A$ and $B,$ since $A^{2}w \in {\rm span}(w,Aw)$ and $A$ has quadratic minimum polynomial. The result now follows by Maschke's Theorem.

2. (Folklore): Let $V$ be a finite dimensional complex vector space, and let $A,B$ be unitary linear transformations on $V$ such that $A,B$ and $AB$ all have order $3$. Then $V$ is an orthogonal direct sum of spaces which are each of dimension $1$ or $3$ and which are all $\langle A,B \rangle$-invariant.

Sketch Proof: Note that $ A^{-1}B^{-1}A^{-1}= (ABA)^{-1} = (BAB)$ since $ABABAB = I.$ Likewise, we have $B^{-1}A^{-1}B^{-1} = ABA$. Then $ H = \langle A^{-1}B, BA^{-1} \rangle$ is Abelian since $AB^{-1}$ and $B^{-1}A$ now commute. Also, $AHA^{-1} \leq \langle BA^{-1},ABA \rangle \leq \langle BA^{-1}, (AB^{-1})(B^{-1}A) \rangle \leq H$ and $A^{-1}HA = \langle ABA, A^{-1}B \rangle = \langle (BA^{-1})^{-1}(A^{-1}B)^{-1}, A^{-1}B \rangle \leq H.$ Hence $A$ normalizes $H$, so that $H \lhd \langle A, B \rangle.$ Note that $\langle A \rangle H = \langle A, B \rangle.$ Thus $H$ has index one or three in $\langle A, B \rangle.$

Now $H$ is an Abelian group of unitary matrices, so may be diagonalized via a unitary matrix. If $w$ is a common eigenvector of length one of each $h \in H,$ then ${\rm span} (w,Aw,A^{2}w)$ is invariant under $\langle A,B \rangle$. Since ${\rm span}(w,Aw,A^{2}w)^{\perp}$ is also invariant under $\langle A,B \rangle,$ we proceed by induction on the dimension of $V$.