Results 1 to 5 of 5

Math Help - Natural embedding

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    70

    Natural embedding

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

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

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Arczi1984 View Post
    How can I show the following:

    If we take natural embedding
    T:A_1 \rightarrow A_1 \oplus A_1
    then it satisfy Txy=TxTy but neither
    \sigma(Tx) \in \sigma(x)
    nor
    Te_1=e_2
    However we have \sigma(x) \in \sigma(Tx).
    To me, the natural embedding of A_1 in A_1 \oplus A_1 would be the map Tx = x\oplus x. But in that case \sigma(Tx) = \sigma(x). Perhaps the question intends the "natural embedding" to be the map 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 e_1 and e_2? If they are identity elements of A_1 and A_2, then Te_1 = e_2 if T is the first of the above maps, but not if T is the second map.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    70
    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:
    Tx = x\oplus 0
    how can I show that
    Te_1=e_2
    where e_1 and e_2 are identity elements of algebras A_1 and A_2 respectively
    and that
    \sigma(x) \in \sigma(Tx)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    3
    You start with  A = C[0,1] and  x(t) = t+1 , then  0 \not\in \sigma(x) but  0 \in \sigma(Tx). Also Te_1 \ne (e_1,e_1) = e_2 , where  e_1 is the constant function x(t) = 1, \forall t \in [0,1]
    Last edited by dineshjk; March 16th 2010 at 11:31 PM. Reason: correcting typos.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    70
    I saw two mistakes (I'm sorry for that).

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

    2. T:A_1 \rightarrow A_1 \oplus A_2 instead of T:A_1 \rightarrow A_1 \oplus A_1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Embedding a group into GL(2, Z)
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 19th 2011, 08:08 AM
  2. Sequence embedding?
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: May 13th 2011, 02:57 AM
  3. Isomorphic embedding
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 1st 2011, 09:35 AM
  4. Embedding of manifolds
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 22nd 2010, 10:57 PM
  5. Sobolev Embedding
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 9th 2010, 02:43 PM

Search Tags


/mathhelpforum @mathhelpforum