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Thread: complex field division algebra?

  1. #1
    Junior Member
    Sep 2008

    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,
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  2. #2
    Global Moderator

    Nov 2005
    New York City
    If $\displaystyle A$ is a finite-dimensional vector space over $\displaystyle \mathbb{C}$ then the map $\displaystyle T:A\to A$ defined by $\displaystyle T(x)=ax$ where $\displaystyle a\in A$ is a linear transformation, that should be easy to check. An eigenvalue is a complex number $\displaystyle k\in \mathbb{C}$ such that there exists a non-zero $\displaystyle x\in A$ so that $\displaystyle T(x) = kx$. Given this linear transformation $\displaystyle T$ we can construct $\displaystyle M=[T]_B$, the matrix representation relative to some basis $\displaystyle B=(b_1,...,b_j)$. Now if we form the polynomial equation* $\displaystyle \det(xI - M) = 0$ we know there is a solution to this equation - since all non-constant polynomial equations have roots in $\displaystyle \mathbb{C}$. Pick any one of them, call it $\displaystyle k$. Then we know there is a non-zero $\displaystyle x\in A$ such that $\displaystyle T(x) = kx \implies ax=kx$. By definition of division algebras we know multiplicative inverse pf $\displaystyle x$ exists. And thus $\displaystyle (ax)x^{-1} = (kx)x^{-1} \implies a = k$. But $\displaystyle k\in \mathbb{C}$ - this implies that $\displaystyle a\in \mathbb{C}$. Thus, $\displaystyle A\subseteq \mathbb{C}\implies A = \mathbb{C}$.

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

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