I am given 3 transformation matrices:
$P = \left( \matrix{0 & 0 & x\\ 0 & 0 & y\\ 0 & 0 & 1} \right)$,
$T_1 = \left( \matrix{s_1\cos(\alpha) & s_2\sin(\alpha) & t_1\\ -s_1\sin(\alpha) & s_2\cos(\alpha) & t_2\\ 0 & 0 & 1} \right)$ such that $-1 < s_1, s_2 < 1$,
$T_2 = \left( \matrix{s_3\cos(\beta) & s_4\sin(\beta) & t_3\\ -s_3\sin(\beta) & s_4\cos(\beta) & t_4\\ 0 & 0 & 1} \right)$ such that $-1 < s_3, s_4 < 1$.
I want to know if an infinite product exists for $P$ such that:
- $P = \overset{\infty}{\underset{i=1}{\prod}}M_i$,
- $M_i \in \{T_1,T_2\}$.
For some specific cases I am able to check for any given $P$ whether a sequence exists or not, but I am really interested in knowing if there is some general rule that can be applied or if something like that is impossible.
A specific example that is easy to solve is when:
$T_1 = \left( \matrix{\frac{1}{2} & 0 & 1\\ 0 & \frac{1}{2} & 0\\ 0 & 0 & 1} \right)$ ,
$T_2 = \left( \matrix{\frac{1}{2} & 0 & 0\\ 0 & \frac{1}{2} & 1\\ 0 & 0 & 1} \right)$.
For this case a sequence exist for $P$ if and only if:
$P = f(z) = \left( \matrix{0 & 0 & z\\ 0 & 0 & 2-z\\ 0 & 0 & 1} \right)$ such that $0 \leq z \leq 2$.
I don't have a very formal proof for this, but I can show that if $P = f(z)$ for $0 \leq z \leq 2$:
$T_1 \cdot P = f(z_1)$ such that $z_1 = \frac{1}{2}z + 1$, which means that $0 \leq 1 \leq z_1 \leq 2$,
$T_2 \cdot P = f(z_2)$ such that $z_2 = \frac{1}{2}z$, which means that $0 \leq z_2 \leq 1 \leq 2$.
We also know that:
${T_1}^\infty = \left( \matrix{0 & 0 & 2\\ 0 & 0 & 0\\ 0 & 0 & 1} \right)$ and
${T_2}^\infty = \left( \matrix{0 & 0 & 0\\ 0 & 0 & 2\\ 0 & 0 & 1} \right)$.
While these statements help prove that:
If for $P$ a sequence, then $P = f(z)$ for $0 \leq z \leq 2$.
To prove it the other way, we can do something similar using $T_1^{-1}$ and $T_2^{-1}$.
If $P = f(z_1)$ for $1 \leq z \leq 2$, then:
- $T_1^{-1} \cdot P = f(z')$ such that $z' = 2z-2$, which means that $0 \leq z' \leq 2$,
If $P = f(z_1)$ for $0 \leq z \leq 1$, then:
- $T_2^{-1} \cdot P = f(z_2)$ such that $z' = 2z$, which means that $0 \leq z' \leq 2$.
Of course the proof is far from sound, but it's only meant as an example. What I'd really like to know is something more general that can be gained out of this.