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1 vote

Finite Parseval Frame

Ok. I have the answer to my own question. It is No. Here is how we prove it: Let $G=\Bbb Z_p^d$. Let $F=\{e_1, e_2, \cdots, e_d\}$ denote the standard orthonormal basis for $\Bbb Z_p^d$. For any $e_i\ …
Melody's user avatar
  • 41
1 vote

Normalized tight frame that is not orthonormal

We call $\{f_n\}$ a normalized tight frame for a Hilbert space $\mathcal H$ if the sequence is a tight frame and all elements have norm one. Now, for any $f_n$ write $$\|f_n\|^2= \sum_m |\langle f_n, …
Melody's user avatar
  • 41
1 vote
1 answer
90 views

Finite Parseval Frame

Assume that $G$ is a finite vector space over a finite field with order $|G|$. (For example, $G=Z_p^k$). Assume that $\{f_n\}_n$ is a Parseval frame for $l^2(G)$. Can we say that the sequence $\{f_n\} …
Melody's user avatar
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1 vote
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What is the analogue of expansive matrix for automorphisms?

We say an invertible $n \times n$ matrix with entries in $\Bbb R^n$ is expansive if the absolute values of all of its eigenvalues exceed $1$. An easy calculation also shows that if we consider a ball …
Melody's user avatar
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