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Math Help - Characteristic polynomial of left-multiplication transformation

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    MHF Contributor Bruno J.'s Avatar
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    Characteristic polynomial of left-multiplication transformation

    Let K be a field, [K:\mathbb{Q}]=n, and take \alpha \in K such that \alpha has degree d over \mathbb{Q}. Let f(x) be the monic irreducible polynomial over \mathbb{Q} having \alpha as a root.

    Let T_\alpha : K \rightarrow K be given by k \mapsto \alpha k. Show that the characteristic polynomial of T_\alpha is f(x)^{n/d}.
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    Quote Originally Posted by Bruno J. View Post
    Let K be a field, [K:\mathbb{Q}]=n, and take \alpha \in K such that \alpha has degree d over \mathbb{Q}. Let f(x) be the monic irreducible polynomial over \mathbb{Q} having \alpha as a root.

    Let T_\alpha : K \rightarrow K be given by k \mapsto \alpha k. Show that the characteristic polynomial of T_\alpha is f(x)^{n/d}.
    from the definition of T_{\alpha}, it's easily seen that for any g(x) \in \mathbb{Q}[x] and k \in K we have g(T_{\alpha})k=g(\alpha)k. thus g(T_{\alpha})=0 if and only if g(\alpha)=0. in particular T_{\alpha} and \alpha have the same minimal

    polynomials. so f(x) is the minimal polynomial of T_{\alpha}. now by, Cayley-Hamilton, the irreducible factors of the characteritsic polynomial and the minimal polynomial of a linear transformation are

    the same. thus the caracteristic polynomial of T_{\alpha} is (f(x))^r, for some integer r \geq 1. comparing the degrees gives us r=\frac{n}{d}.
    Last edited by NonCommAlg; June 16th 2009 at 03:59 PM.
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