Prove $\frac{Ax}{ \Vert Ax \Vert}$ is a contraction mapping I am going through a proof of the Perron-Frobenius theorem which uses the Banach fixed point theorem. The author first asks the reader to consider the space
$$
X = \left\{ x \in  \mathbb{R}^{d} : \Vert x\Vert^{2} = 1 \text{ and } x_{i} \geq 0 \right\}
$$
Then consider the matrix $A$ with strictly positive entries. We define the map:
$$
T(x) = \frac{Ax}{\Vert Ax \Vert}
$$
The authors then argue that, since T is a contraction mapping with respect to the geodesic sphere distance, the Banach fixed point Theorem can be applied. This results in a unique fixed point satisfying the conditions of the Theorem. That is, there exists only one eigenvector with strictly positive entries (Proving that the corresponding eigenvalue is maximal is done in the latter part of the proof).
I understand the overall idea of the proof but am struggling to work out why $T$ is a contraction mapping. The spherical distance is given by:
$$
d(x, y) = \cos^{-1}(x^{\top} y)
$$
Thus we need to prove that
$$
\cos^{-1}\left(\frac{x^{\top}A^{T}Ay}{\Vert Ax\Vert \Vert Ay \Vert}\right) \leq k\cos^{-1}(x^\top y)
$$
for some $0 < k < 1$. One observation I tried using is that $\cos^{-1}$ monotonically decreasing on the domain $[-1, 1]$. Thus being able to show
$$
\cos^{-1}\left(\frac{x^{\top}A^{T}Ay}{\Vert Ax\Vert \Vert Ay \Vert}\right) \leq \cos^{-1}(x^\top y)
$$
is equivalent to showing that
$$
\frac{x^{\top}A^{T}Ay}{\Vert Ax\Vert \Vert Ay \Vert} \geq x^\top y
$$
My plan was then to use basic facts about operator norms to prove this inequality but I couldn't make any progress.
EDIT: This question is based on the following lecture notes.
 A: The following is a counterexample.
$$
A = \begin{pmatrix}
    1.00 &    0.01 &    0.01 \\
    0.01 &    1.00 &    0.01 \\
    0.01 &    0.01 &    0.01 \\
\end{pmatrix},
\quad x =
\begin{pmatrix}
    0.4748 \\
    0.0028 \\
    0.8801 \\
\end{pmatrix},
\quad y =
\begin{pmatrix}
    0.0028 \\
    0.4748 \\
    0.8801 \\
\end{pmatrix}.
$$
Then,
$$
\frac{x^\top A^\top A y}{\|Ax\|\,\|Ay\|} = 0.0683 < 0.7772 = x^\top y.
$$
A: Rewriting,
\begin{align*}
(Ax)^\intercal(Ay) \ge (x^\intercal y)\|Ax\|\|Ay\|
\end{align*}
Note that, since all elements of $A$ are positive and elements of $x, y \in X$ are nonnegative, both sides of the equality are positive. Squaring retains the $\ge$ inequality, and using the fact that $\|x\|^2 = x^\intercal x$, we have
\begin{align*}
(x^\intercal A^\intercal A x)(y^\intercal A^\intercal A y) \ge (x^\intercal y)^2(x^\intercal A^\intercal A x)(y^\intercal A^\intercal A y)
\end{align*}
And again, because of the positivity (non-negativity) of the elements of $A$ ($x, y$, respectively), $x^\intercal A^\intercal A x$ and $y^\intercal A^\intercal A y$ are all $\ge 0$, so we can safely divide them from both sides to arrive at
\begin{align*}
1 \ge (x^\intercal y)^2
\end{align*}
which is true since $\|x\| = \|y\| = 1$ and Cauchy-Schwarz.
