If is a finite-dimensional vector space over then the map defined by where is a linear transformation, that should be easy to check. An eigenvalue is a complex number such that there exists a non-zero so that . Given this linear transformation we can construct , the matrix representation relative to some basis . Now if we form the polynomial equation* we know there is a solution to this equation - since all non-constant polynomial equations have roots in . Pick any one of them, call it . Then we know there is a non-zero such that . By definition of division algebras we know multiplicative inverse pf exists. And thus . But - this implies that . Thus, .
If and is odd then by repeating the same argument we arrive at the polynomial equation . However the degree of this polynomial equation is which is odd - and every odd degree polynomial over has a real root. Use this infromation to show .
*)Remember from linear algebra that this is the polynomial we need to solve to find all eigenvalues.