Uniform Hyperbolicity Decay Estimate This question has been posted on math overflow with little interest, so I am posting it here.  https://mathoverflow.net/questions/139010/uniform-hyperbolicity-decay-estimate
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 (Inspired by A. Blumenthal's comment):  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. 
 A: No need to fudge with $\lambda, B$ to make it work. We've overlooked the fact that when $\|\cdot \|$ is the matrix norm induced by the euclidean norm on $\mathbb{R}^2$, then for $A \in SL_2(\mathbb{R})$, $$ \|A\| = \|A^{-1}\| $$.
As Alfred notes, this is false
This follows by examining the singular value decomposition carefully for $A^{-1}$, and noting that the singular values for $A$ are always $\{\|A\|,\|A\|^{-1}\}$.
Now use that $A_{n+k}(x) =A_k(T^nx) \circ A_n(x)$ to obtain
$$
\|A_{n+k}(x)\|^{-1} = \|\big( A_{n+k}(x) \big) ^{-1}\| \leq \|\big( A_{n}(x) \big) ^{-1} \| \|\big( A_{k}(T^n x) \big) ^{-1}\| = \|A_n(x)\|^{-1} \cdot \|A_k(T^nx)\|^{-1}
$$
This, in conjunction with what you've already shown, will complete the exponential decay estimate.

Update (2/25/2017)
Bochi and Gourmelon give an proof essentially extending Yoccoz's method to a related result for general matrix cocycles. I also wrote a paper (with a completely different proof) with Ian Morris extending this result to the setting of Banach space cocycles.
