Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem: Let $\{A_r\},\{B_r\},\{C_r\}$ each be families of (determinant 1) 2x2 matrices in $SL(2,\mathbb{R})$ such that each family is continuously indexed by a parameter $0<r<1$. Is there a known formula for calculating $tr(W)$, where $W$ is $(A_rB_rC_r)^nA_r$, $(A_rB_rC_r)^nA_rB_r$, or $(A_rB_rC_r)^n$?

Progress thus far: The indexing is sufficiently complicated that simply multiplying and looking for a pattern has not been successful. Each entry of each family is a ratio of quadratics in $r$, and I do not remember the right computational linear algebra to attack this problem.

Motivation: the families of matrices represent isometries of the hyperbolic plane, and I am interested in knowing when such isometries change from elliptic to parabolic to hyperbolic

share|cite|improve this question
I've adjusted the question to say that I'm looking at matrices with determinant 1. I only care about the absolute value of the trace, but I'm looking to plot it as a function of the indexing variable $r$, so a polynomial output would be nice. – KReiser Aug 1 '12 at 2:07
If you can solve for the eigenvalues and eigenvectors of $A_r B_r C_r$ as a function of $r$ (this is something of a big if), then you can work over $\mathbb{C}$ and change basis so that $A_r B_r C_r$ is in Jordan normal form. Things are not so bad from here. – Qiaochu Yuan Aug 1 '12 at 2:22
Instead of thinking of them as families of matrices in $SL(2,\mathbb R)$, you can instead think of them as single matrices $A,B,C \in SL\big(2,{\mathbb R}(r)\big)$. – Greg Martin Aug 1 '12 at 5:53
up vote 4 down vote accepted

$\def\Tr{\mathrm{Tr}}$Let $M(r) = A(r) B(r) C(r)$. Let the characteristic polynomial of $M(r)$ be $$\lambda^3 + x(r) \lambda^2 + y(r) \lambda + z(r)$$ Each of $x(r)$, $y(r)$ and $z(r)$ is a rational function, computable by a computer algebra system.

If you can solve this cubic then proceed as Qiaochu says. (And it is definitely worth putting this cubic into a computer algebra system to see whether it simplifies in some surprising way.) But solving cubics is usually painful, so here is an alternative:

The Cayley-Hamilton theorem tells you that $$M(r)^3 + x(r) M(r)^2 + y(r) M(r) + z(r)=0$$ so $$\Tr M(r)^{n+3} = - x(r) \Tr M(r)^{n+2} - y(r) \Tr M(r)^{n+1} - z(r) \Tr M(r)^n$$ So this gives a linear recursion for $\Tr M(r)^n$ which you can use to compute $\Tr M(r)^n$ pretty easily, and similarly for your other traces.

share|cite|improve this answer
Do you think this method could work for this question:… – Jp McCarthy Nov 9 '15 at 9:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.