# Thread: complex field division algebra?

1. ## 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

2. 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.