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Math Help - Normal linear maps in polynomials

  1. #1
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    Normal linear maps in polynomials

    I am trying to solve a problem and I need to know whether I can deduce the following:

    Suppose T is normal, and there exist a polynomial P(t) such that P(T)v = 0. Then <P(T)u,v> = <u,P(T)^{*}v> = <u,0> = 0. The part I am not so sure about is that P(T)^{*}v=0.

    Can I say that and if so why?

    Any direction will be appreciated

    SK
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by skyking View Post
    I am trying to solve a problem and I need to know whether I can deduce the following:

    Suppose T is normal, and there exist a polynomial P(t) such that P(T)v = 0. Then <P(T)u,v> = <u,P(T)^{*}v> = <u,0> = 0. The part I am not so sure about is that P(T)^{*}v=0.

    Can I say that and if so why?

    Any direction will be appreciated

    SK
    How about a rank argument?
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  3. #3
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    Can't think of a rank argument. How would you show this using ranks?

    I was able to figure it out though, here's my proof.

    Let's say S=P(T), since T is normal it follows S is normal, therfore S can be represented by a diagonal matrix D=diag{(d_{1},...,d_{n})}. We know Sv=0, that means that Dv=0 and therfore (d_{1}a_{1},...,d_{n}a_{n})=0 where a_{i} are the coordinates of v. So for every 1\leq i \leq n we get d_{i}a_{i}=0. Now if we look at S^{*} it can be representd by D^{*} and therfore S^{*}v=D^{*}v=(\bar{d_{1}}a_{1},...\bar{d_{n}}a_{n  })=0
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  4. #4
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    Thinking about this again, I am not so sure that I can say that if T is normal then P(T) is normal. Can I say this?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by skyking View Post
    Thinking about this again, I am not so sure that I can say that if T is normal then P(T) is normal. Can I say this?
    Let \displaystyle P(T)=\sum_{j=0}^{n}\alpha_j T^j we then have that \displaystyle P(T)^\ast=\sum_{j=0}^{n}\overline{\alpha_j}\left(T  ^j\right)^\ast. Our first claim that T^k\left(T^j\right)^\ast=\left(T^j\right)^\ast T^k for every k,j\in\mathbb{N}\cup\{0\}. We do this by letting k be arbitrary and proceeding by induction on j. Evidently this is true when j=0 and so assume it's true for j then

    \begin{aligned}T^k\left(T^{j+1}\right)^\ast &=\left(T^k \left(T^j\right)^\ast T^\ast\\ &=\left(T^j\right)^ast T^k T^\ast\\ &=\left(T^j\right)^\ast T^{k-1}TT^\ast\\ &= \left(T^j\right)^\ast T^{k-1}T^\ast T\\ &= \vdots\\ &= \left(T^j\right)^\ast T^\ast T^k\\ &= \left(T^{j+1}\right)^\ast T^k\end{aligned}


    then with this we see that


    \displaystyle \begin{aligned}P(T)P(T)^\ast &= \left(\sum_{k=0}^{n}\alpha_k T^k\right)\left(\sum_{j=0}^{n}\overline{\alpha_j}\  left(T^j\right)^\ast\right)\\ &= \sum_{k=0}^{n}\sum_{j=0}^{n}\left(\alpha_j T^k\right)\left(\overline{\alpha_j}\left(T^\ast\ri  ght)^j\right)\\ &= \sum_{k=0}^{n}\sum_{j=0}^{n}\left(\overline{\alpha  _j}\left(T^j\right)^\ast\right)\left(\alpha_k T^k\right)\\ &= \left(\sum_{j=0}^{n}\overline{\alpha_j}\left(T^j\r  ight)^\ast\right)\left(\sum_{k=0}^{n}\alpha_k T^k\right)\\ &= P(T)^\ast P(T)\end{aligned}
    Last edited by Drexel28; January 16th 2011 at 08:10 PM.
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