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Math Help - Operator has no eigenvalues

  1. #1
    Senior Member
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    Operator has no eigenvalues

    Hello everyone

    Show: The operator T has no eigenvalues

    T:l^2(\mathbb{C}) \to l^2(\mathbb{C})

    Tx:=\sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1} , where x = (x_k)_{k \in \mathbb{N}}

    Proof:

    Iff \lambda \not= 0 it follows that x_1 = 0 and x_{k+1} = \frac{x_k}{\lambda(k+1)} if  k \ge 1

    \Rightarrow x = 0

    I don't get it: how do we know that x_1 = 0 ?

    Help would be much appreciated

    Thanks
    Rapha
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  2. #2
    Member mabruka's Avatar
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    Mexico City
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    Hi


    I think it should say (Tx)_1=0.

    You agree?

    It is implicit in the definition of T that



    Tx = 0\cdot e_1  + \sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Rapha View Post
    Hello everyone

    Show: The operator T has no eigenvalues

    T:l^2(\mathbb{C}) \to l^2(\mathbb{C})

    Tx:=\sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1} , where x = (x_k)_{k \in \mathbb{N}}

    Proof:

    Iff \lambda \not= 0 it follows that x_1 = 0 and x_{k+1} = \frac{x_k}{\lambda(k+1)} if  k \ge 1

    \Rightarrow x = 0

    I don't get it: how do we know that x_1 = 0 ?
    Look at the definition of T- the sum is \frac{x_1}{2}e_2+ \frac{x_2}{3}e_3+ \cdot\cdot\cdot. The crucial point is that there is NO " e_1" in that. In order for that to be a multiple of x, there must be NO " e_1" in x. That means that the coefficient of e_1, x_1, must be 0.

    Help would be much appreciated

    Thanks
    Rapha
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  4. #4
    Senior Member
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    Hi!

    Thank you so much! It would never have occurred to me. Your explanations made this excercise so easy.

    Rapha
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