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

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

    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 $\displaystyle S=P(T)$, since $\displaystyle T$ is normal it follows $\displaystyle S$ is normal, therfore S can be represented by a diagonal matrix $\displaystyle D=diag{(d_{1},...,d_{n})}$. We know $\displaystyle Sv=0$, that means that $\displaystyle Dv=0$ and therfore $\displaystyle (d_{1}a_{1},...,d_{n}a_{n})=0$ where $\displaystyle a_{i}$ are the coordinates of $\displaystyle v$. So for every $\displaystyle 1\leq i \leq n$ we get $\displaystyle d_{i}a_{i}=0$. Now if we look at $\displaystyle S^{*}$ it can be representd by $\displaystyle D^{*}$ and therfore $\displaystyle S^{*}v=D^{*}v=(\bar{d_{1}}a_{1},...\bar{d_{n}}a_{n })=0$
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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 $\displaystyle \displaystyle P(T)=\sum_{j=0}^{n}\alpha_j T^j$ we then have that $\displaystyle \displaystyle P(T)^\ast=\sum_{j=0}^{n}\overline{\alpha_j}\left(T ^j\right)^\ast$. Our first claim that $\displaystyle T^k\left(T^j\right)^\ast=\left(T^j\right)^\ast T^k$ for every $\displaystyle k,j\in\mathbb{N}\cup\{0\}$. We do this by letting $\displaystyle k$ be arbitrary and proceeding by induction on $\displaystyle j$. Evidently this is true when $\displaystyle j=0$ and so assume it's true for $\displaystyle j$ then

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


    then with this we see that


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