I have been trying to understand the proof of the following result, which is considered well-known.

Theorem: Fix a compact metric space $X$, a homeomorphism $T:X \to X$, and a continuous map $ A : X \to \mathrm{SL}_2(\mathbb{R}) $. Define the skew-product $$ (T,A): X \times \mathbb{R}^2 \to X \times \mathbb{R}^2, \quad (x,v) \mapsto (Tx, A(x)v), $$ and set $ A_n(x) = A(T^{n-1}x) \cdots A(x) $ for $x \in X, n > 0$ and similarly for $n \leq 0 $ so that $ (T,A)^n = (T^n,A_n) $. If there are uniform constants $ C > 0, \lambda > 1 $ such that $$ \| A_n(x) \| \geq C \lambda^{|n|} $$ for all $n,x$, then the cocycle $(T,A)$ is uniformly hyperbolic. More precisely, there are continuous maps $ \Lambda^s,\Lambda^u: X \to \mathbb{RP}^1 $ and constants $ c >0 , L>1$ such that $$ A(x) \Lambda^{\bullet}(x) = \Lambda^{\bullet}(Tx), \quad \bullet \in \{ s,u \} $$ and $$ \| A_n(x) v_s \| \leq cL^{-n} \|v_s\|, \quad \| A_{-n}(x) v_u \| \leq cL^{-n} \| v_u \| $$ for all $n \geq 0, x \in X, v_s \in \Lambda^s(x), v_u \in \Lambda^u(x)$.

The consruction of $\Lambda^s$ goes like this: Given $ x \in X, n >0 $, let $ \Lambda_n^s(x) $ be the most contracted subspace of $ A_n(x) $ (i.e. the eigenspace of $ A_n(x)^* A_n(x) $ corresponding to the eigenvalue $ \|A_n(x) \|^{-2} $). Of course, one needs $ \| A_n(x) \| > 1 $ for this subspace to be one-dimensional, but the growth condition assures us that this happens for sufficiently large $n$.

One then proves readily that there are constants $C_0,C_1$ independent of $x$ and $n$ such that the angles between these singular subspaces obey $$ \angle \left( \Lambda_n^s(x), \Lambda_{n+1}^s(x) \right) \leq C_0 \| A_n(x) \|^{-2} \leq C_1 \lambda^{-2n}. $$ In particular, $ \Lambda_n^s(\cdot) $ converges (uniformly) to a limiting map $ \Lambda^s(\cdot) $. Continuity of $\Lambda^s$ is immediate, and the $A$-invariance condition follows from a straightforward calculation.

Now, here is the part of the proof with which I am having difficulties - the exponential decay estimates. Pick a unit vector $v_s \in \Lambda^s(x)$, and let $ \theta_n = \theta_n(x) $ denote the (smallest nonnegative) angle between $\Lambda_n^s(x)$ and $\Lambda^s(x)$. One can check that $$ \| A_n(x) v_s \|^2 = \| A_n(x) \|^{-2} \cos^2(\theta_n) + \| A_n(x) \|^{2} \sin^2(\theta_n) $$ (simply decompose $v_s$ in an orthonormal basis consisting of a unit vector from $\Lambda_n^s(x)$ and a unit vector from $\Lambda_n^u(x)$). We want to see that this decays exponentially. That the first term on the RHS decays exponentially is obvious, but the second term is bothersome. The factor $ \| A_n(x) \|^2 $ grows exponentially, and the factor $ \sin^2(\theta_n) $ decays exponentially, but it is not clear to me why the exponential decay of $\sin^2(\theta_n)$ should necessarily ``win'' and produce a net result of exponential decay.

I have a nice reference for this result, namely Yoccoz' article ``Some questions and remarks about $ \mathrm{SL}_2(\mathbb{R}) $ cocycles.'' Unfortunately for me, the decay estimate I want to understand is referred to as something easily checked, so I am likely missing something very obvious. I would be very grateful for any helpful remarks.

EDIT: Due to lack of interest, I have cross-posted this question on math stack exchange: https://math.stackexchange.com/questions/467478/uniform-hyperbolicity-decay-estimate

EDIT (Inspired by A. Blumenthal's comment on Math Stack Exchange): The bound on the angle between $ \Lambda_n $ and $\Lambda_{n+1}$ implies that $$ \theta_n(x) \leq C_0 \sum_{m=n}^\infty \| A_m(x) \|^{-2}, $$ so it would be enough to prove that $$ \| A_n(x) \|^2 \left( \sum_{m=n}^\infty \| A_m(x) \|^{-2} \right)^2, $$ decays exponentially. If we define $ a_n(x) = \| A_n(x) \| $ and $ B = \sup_{x\in X} \|A(x) \| $, then we have a sequence of functions $ a_n:X \to \mathbb{R}_{\geq 0} $ with the following properties:

$\bullet$ $ C \lambda^n \leq a_n(x) \leq B^n $

$\bullet$ $ B^{-1} \cdot a_n(x) \leq a_{n+1}(x) \leq B\cdot a_n(x) $

$\bullet$ $ B^{-2} \cdot a_n(x) \leq a_n(Tx) \leq B^2 \cdot a_n(x) $

for all $x \in X$, $n \geq 0$.

We would then like to show that $$ a_n(x) \sum_{m=n}^\infty a_m(x)^{-2} $$ decays exponentially (uniformly in $x$). This looks more promising, but the desired estimate remains elusive.

In particular, if the sequence $a_n(x)^{-2}$ decreases monotonically, or if $ B<\lambda^2 $, the desired estimate is obvious. However, if $B $ is much larger than $\lambda^2$ and the sequence is wildly non-monotonic, then the waters remain murky to me.