設(b$_{i,j}$)是一個定義在 extit{ l}$^{2}$上的有界矩陣，$BbbT={zinBbb C:|z|=1}$，A是一個在L$^{2}(mathbb{T)}$上的有界矩陣滿足下列情形
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 $

對於所有$i,jin mathbb{Z}$, 其中$sigma =1+|alpha
|^{2},,alpha in mathbb{C},alpha
eq0$

上述情形給了我們一個二冪次遞迴關係。下圖表示矩陣第$ij$項如何生成對應的二乘二的區塊的項 {$a_{2i,2j}, a_{2i-1,2j}, a_{2i,2j-1}, a_{2i-1,2j-1}$ }

egin{figure}[hp]
egin{center}
includegraphics[scale=0.42]{cubic.pdf}
end{center}
caption{此二幕次遞迴的形式} end{figure}
由於[2]中可知$displaystyle A=sum_{n=0}^{infty }S^{n}BS^{ast
n}$, 其中
$ Sz^i=sigma ^{-1/2}(overline{alpha }z^{2i}+z^{2i-1})$

$ B=sum sum b_{ij}(u_{i}otimes u_{j})$ ;;; which
$u_{i}(z)=sigma ^{-1/2}z^{2i-1}(alpha -z)$
則我們可以用它來求出符合上述情形的$a_{ij}$的明確公式。

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

1 Introduction -------------------------------------iv
2 The operators that constant with S ------vii

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