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

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!

• Fixed typo in title. – David E Speyer Apr 21 at 11:45

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.

1. (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 $$V$$.

• Thanks, this second generalization is interesting for me! Is it possible to generalize further to higher number (not just n=2 or 3)? – MarcO Apr 21 at 20:35
• It becomes more difficult. It is necessary to know some representation theory to answer more carefully. . If you have two unitary matrices which generate a finite group, there are various theorems of Burnside, Frobenius and Schur which are sometimes helpful. However, I could not think of too many general statements of interest wwhich were reasonably clean. – Geoff Robinson Apr 21 at 21:29