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Thread: Natural embedding

  1. #1
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    Natural embedding

    Hi,
    This is the next problem in that theme. How can I show the following:

    If we take natural embedding
    $\displaystyle T:A_1 \rightarrow A_1 \oplus A_1$
    then it satisfy $\displaystyle Txy=TxTy$ but neither
    $\displaystyle \sigma(Tx) \in \sigma(x)$
    nor
    $\displaystyle Te_1=e_2$
    However we have $\displaystyle \sigma(x) \in \sigma(Tx)$.

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    How can I show the following:

    If we take natural embedding
    $\displaystyle T:A_1 \rightarrow A_1 \oplus A_1$
    then it satisfy $\displaystyle Txy=TxTy$ but neither
    $\displaystyle \sigma(Tx) \in \sigma(x)$
    nor
    $\displaystyle Te_1=e_2$
    However we have $\displaystyle \sigma(x) \in \sigma(Tx)$.
    To me, the natural embedding of $\displaystyle A_1$ in $\displaystyle A_1 \oplus A_1$ would be the map $\displaystyle Tx = x\oplus x$. But in that case $\displaystyle \sigma(Tx) = \sigma(x)$. Perhaps the question intends the "natural embedding" to be the map $\displaystyle Tx = x\oplus 0$? In that case, 0 is always in the spectrum of Tx, but it need not be in the spectrum of x.

    What are $\displaystyle e_1$ and $\displaystyle e_2$? If they are identity elements of $\displaystyle A_1$ and $\displaystyle A_2$, then $\displaystyle Te_1 = e_2$ if T is the first of the above maps, but not if T is the second map.
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  3. #3
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    In that paper it is called 'natural embedding' but they didn't give any formula. When I was trying to solve this problem I took the second case witch You are mentioned.

    How can I show that o need not be in the spectrum of x (in this case)?
    If we take this natural embedding:
    $\displaystyle Tx = x\oplus 0$
    how can I show that
    $\displaystyle Te_1=e_2$
    where $\displaystyle e_1$ and $\displaystyle e_2$ are identity elements of algebras $\displaystyle A_1$ and $\displaystyle A_2$ respectively
    and that
    $\displaystyle \sigma(x) \in \sigma(Tx)$?
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  4. #4
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    You start with $\displaystyle A = C[0,1]$ and $\displaystyle x(t) = t+1 $, then $\displaystyle 0 \not\in \sigma(x)$ but $\displaystyle 0 \in \sigma(Tx)$. Also $\displaystyle Te_1 \ne (e_1,e_1) = e_2 $, where $\displaystyle e_1 is the constant function x(t) = 1, \forall t \in [0,1]$
    Last edited by dineshjk; Mar 16th 2010 at 11:31 PM. Reason: correcting typos.
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  5. #5
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    I saw two mistakes (I'm sorry for that).

    1. Of course it shoul be: $\displaystyle \sigma(Tx) \subset \sigma(x)$ and $\displaystyle \sigma(x) \subset \sigma(Tx)$

    2. $\displaystyle T:A_1 \rightarrow A_1 \oplus A_2$ instead of $\displaystyle T:A_1 \rightarrow A_1 \oplus A_1$
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