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Math Help - complex field division algebra?

  1. #1
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    complex field division algebra?


    I have to hand this in tomorrow and I'll have to go to bed soon. I will have 3hours between classes tomorrow to work on it as well. If someone could show me which direction I should go in or give me some hints that would be much appreciated.
    Thanks you,
    Jean-Bernard
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  2. #2
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    If A is a finite-dimensional vector space over \mathbb{C} then the map T:A\to A defined by T(x)=ax where a\in A is a linear transformation, that should be easy to check. An eigenvalue is a complex number k\in \mathbb{C} such that there exists a non-zero x\in A so that T(x) = kx. Given this linear transformation T we can construct M=[T]_B, the matrix representation relative to some basis B=(b_1,...,b_j). Now if we form the polynomial equation* \det(xI - M) = 0 we know there is a solution to this equation - since all non-constant polynomial equations have roots in \mathbb{C}. Pick any one of them, call it k. Then we know there is a non-zero x\in A such that T(x) = kx \implies ax=kx. By definition of division algebras we know multiplicative inverse pf x exists. And thus (ax)x^{-1} = (kx)x^{-1} \implies a = k. But k\in \mathbb{C} - this implies that a\in \mathbb{C}. Thus, A\subseteq \mathbb{C}\implies A = \mathbb{C}.

    If F = \mathbb{R} and \text{dim}(A) is odd then by repeating the same argument we arrive at the polynomial equation \det( xI - M) = 0. However the degree of this polynomial equation is \text{dim}(A) which is odd - and every odd degree polynomial over \mathbb{R} has a real root. Use this infromation to show A = \mathbb{R}.

    *)Remember from linear algebra that this is the polynomial we need to solve to find all eigenvalues.
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