Frobenius theorem on real division algebras

Question

Can someone help me with this? I don't understand conclusion after Exercise 7, that A is direct sum of eigenspaces U(1) and U(-1). Thanks in advance :) http://danshved.wordpress.com/2013/05/09/frobenius-theorem-on-real-division-algebras/

Answer 1

Since the invertible linear map $\phi$ satisfies the equation $X^2-1=0$, the minimal polynomial of $\phi$ divides $x^2-1$. Hence $\phi$ is diagonalizable, and has eigenvalues $1$ and $-1$. Hence there is a basis of eigenvectors, i.e., $U(1)\oplus U(-1)=A$. For details see here.