Description
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Let $\{e\_n\}$, $e\_n\in \mathbb{R}^p$ be a sequence of vectors, $\{U\_n\}$, $U\_n\in\mathbb{C}^{p\times p}$ be a sequence of **unitary matrices** (that is $U\_i^\*=U\_i^{-1}$, $^\*$denonts conjugate transpose). $Q\in\mathbb{R}^{p\times p}$ is a diagonal matrix with positive diagonal entries. They satisfy the following properties:


$||QU\_1e\_2||\_2<\rho||QU\_1e\_1||\_2$,


$||QU\_2e\_3||\_2<\rho||QU\_2e\_2||\_2$,


$||QU\_3e\_4||\_2<\rho||QU\_3e\_3||\_2$,


...........


$||QU\_ie\_{i+1}||\_2<\rho||QU\_ie\_i||\_2$,...., where $\rho\in(0,1)$, $||\cdot||\_2$ denotes $l\_2$ norm on vectors.


Questions
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**Does $\{e\_n\}$ converge to zero?** 


Some observations: consider a special case where $U\_i=I$, then $e\_n\rightarrow 0$. That means $\{e\_n\}$ converges to zero under certain condition. I wonder if it converges when $\{U\_n\}$ is an arbitrary sequence of unitary matrices.



Does this sequence converge to zero?
Here is a counterexample: Let $p=2$ and let $Q$ be the diagonal matrix with entries $(2,1)$. If $n$ is odd, let $e\_n$ be the first standard basis vector $(1,0)^t$ and if $n$ is even, the second $(0,1)^t$.
If $n$ is odd, let $U\_n$ be the identity matrix and if $n$ is even, let 
$U\_n$ be the matrix with first row (0,1) and second row (1,0), so $U\_n$ just interchanges the two standard basis vectors.


