Title page for etd-0725107-195949


[Back to Results | New Search]

URN etd-0725107-195949
Author Chia-ming Hsu
Author's Email Address lozf1024@yahoo.com.tw
Statistics This thesis had been viewed 5078 times. Download 0 times.
Department Applied Mathematics
Year 2006
Semester 2
Degree Master
Type of Document
Language English
Title On infinite matrices whose entries satisfying certain dyadic recurrent formula
Date of Defense 2007-06-15
Page Count 22
Keyword
  • slant Toeplitz operator
  • shift operator
  • separable Hilbert space
  • bounded matrix
  • dyadic recurrent formula
  • Abstract Let (b$_{i,j}$) be a bounded matrix on extit{ l}$^{2}$, $Bbb
    T={zinBbb C:|z|=1}$, and A be a bounded matrix on L$^{
    2}(mathbb{T)}$ satisfying the conditions
     1.$langle Az^{2j},z^{2i}
    angle =sigma ^{-1}b_{ij}+|alpha
    |^{2}sigma ^{-1}langle Az^{j},z^{i}
    angle $;
     2.$langle Az^{2j},z^{2i-1}
    angle =-alpha sigma
    ^{-1}b_{ij}+alpha sigma ^{-1}langle Az^{j},z^{i}
    angle $;
     3.$langle Az^{2j-1},z^{2i}
    angle =-overline{alpha }sigma
    ^{-1}b_{ij}+overline{alpha }sigma ^{-1}langle
    Az^{j},z^{i}
    angle$;
     4.$langle Az^{2j-1},z^{2i-1}
    angle =|alpha |^{2}sigma
    ^{-1}b_{ij}+sigma ^{-1}langle Az^{j},z^{i}
    angle $
    hspace{-0.76cm} for all $i,jin mathbb{Z}$, where $sigma
    =1+|alpha |^{2},,alpha in mathbb{C},alpha
    eq0$.
    The above conditions evidently suggests that there is a "dyadic"
    relation in the entries of $A$. Here in the following picture
    illustrates how each $ij-$th entry of $A$ generates the 2 by 2 block
    in $A$ with entries ${a_{2i 2j}, a_{2i-1 2j}, a_{2i 2j-1},
    a_{2i-1 2j-1}}.$ vspace{-0.3cm}
    egin{figure}[hp]
    egin{center}
    includegraphics[scale=0.42]{cubic.pdf}
    end{center}
    vspace{-0.8cm}caption{The dyadic recurrent form} end{figure}
    It has been shown [2] that $displaystyle A=sum_{n=0}^{infty
    }S^{n}BS^{ast n}$, where $Sz^i=sigma ^{-1/2}(overline{alpha
    }z^{2i}+z^{2i-1})$ and $$B=sumlimits_{i=-infty}^infty
    sumlimits_{j=-infty}^infty b_{ij}(u_{i}otimes u_{j}),
    u_{i}(z)=sigma ^{-1/2}z^{2i-1}(alpha -z).$$
    In this paper, we shall use the above relations to compute $langle
    a_{i,j}
    angle $ explicitly.
    ewline
    Key words: shift operator, bounded matrix, dyadic recurrent formula,
    slant Toeplitz operator, separable Hilbert space
     2.$langle Az^{2j},z^{2i-1}
    angle =-alpha sigma
    ^{-1}b_{ij}+alpha sigma ^{-1}langle Az^{j},z^{i}
    angle $
     3.$langle Az^{2j-1},z^{2i}
    angle =-overline{alpha }sigma
    ^{-1}b_{ij}+overline{alpha }sigma ^{-1}langle
    Az^{j},z^{i}
    angle $
     4.$langle Az^{2j-1},z^{2i-1}
    angle =|alpha |^{2}sigma
    ^{-1}b_{ij}+sigma ^{-1}langle Az^{j},z^{i}
    angle $
     for all $i,jin mathbb{Z}$, where $sigma =1+|alpha
    |^{2},,alpha in mathbb{C},alpha
    eq0$
    egin{figure}[hp]
    egin{center}
    includegraphics[scale=0.42]{cubic.pdf}
    end{center}
    caption{The dyadic recurrent form} end{figure}
    Since it has been
    shown [2] that $displaystyle A=sum_{n=0}^{infty }S^{n}BS^{ast
    n}$, where
    $   Sz^i=sigma ^{-1/2}(overline{alpha }z^{2i}+z^{2i-1})$
    $   B=sum sum b_{ij}(u_{i}otimes u_{j})$ ;;; which
    $u_{i}(z)=sigma ^{-1/2}z^{2i-1}(alpha -z)$
    Then we can use it to compute $langle Az^{j},z^{i}
    angle $
    explicity if A satisfies the previous condition.
    ewline
    Key words: shift operator, bounded matrix, dyadic recurrent formula,
    slant Toeplitz operator, separable Hilbert space
    Advisory Committee
  • Mu ming Wong - chair
  • Jyh-Shyang Jeang - co-chair
  • Ngai-Ching Wong - co-chair
  • Mark C. Ho - advisor
  • Files
  • etd-0725107-195949.pdf
  • indicate not accessible
    Date of Submission 2007-07-25

    [Back to Results | New Search]


    Browse | Search All Available ETDs

    If you have more questions or technical problems, please contact eThesys