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Math Help - Minimal polynomial problem

  1. #1
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    Minimal polynomial problem

    Prove that if T is invertible and the degree of  minP_{T} = m
    then
     minP_{T^{-1}}(x) = cx^{m}minP_{T}(x^{-1})

    where  c=minP_{T}(0)^{-1}
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  2. #2
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    Regardless of getting the actual proof of this problem, I am simply just having trouble seeing why this would be true in the first place.
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  3. #3
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    Quote Originally Posted by linalguser3 View Post
    Prove that if T is invertible and the degree of  minP_{T} = m
    then
     minP_{T^{-1}}(x) = cx^{m}minP_{T}(x^{-1})

    where  c=minP_{T}(0)^{-1}
    Fact: suppose p(x)=x^m + c_{m-1}x^{m-1} + \cdots + c_1x + c_0 is irreducible over some field. then q(x)=c_0^{-1}x^m p(x^{-1})=x^m + c_1c_0^{-1}x^{m-1} + \cdots + c_{m-1}c_0^{-1}x + c_0^{-1} is irreducible too.

    Proof: suppose q(x)=(x^r + a_{r-1}x^{r-1} + \cdots + a_0)(x^s + b_{s-1}x^{s-1} + \cdots + b_0). then c_0^{-1}x^{-m}p(x)=q(x^{-1})=(x^{-r} + a_{r-1}x^{-r+1}+ \cdots + a_0)(x^{-s} + b_{s-1}x^{-s+1} + \cdots + b_0). therefore:

    p(x)=(x^r + a_1a_0^{-1}x^{r-1} + \cdots + a_0^{-1})(x^s + b_1b_0^{-1}x^{s-1} + \cdots + b_0^{-1}), which is impossible because p(x) is irreducible.

    using what i just proved, you should be able to solve your problem quite easily!
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