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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

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.

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1 Answer 1

up vote 1 down vote accepted

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.

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Great! Thanks a lot. I only want to pint out a typo. If $n$ is even, $e_n=(0,1)^t$. –  Zhang Changhe Sep 21 '12 at 7:48
    
Thanks. Changed it. –  doug Sep 21 '12 at 11:16
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