Let p be an endomorphism of some vector space V over K. Show: If all vectors of V are eigenvectors for p (to possibly different eigenvalues), then p is a homothety (ie p=r(id) for some r element of K).
You're given that every vector is an eigenvector of P and want to show that all the eigenvalues are the same. Suppose not, that is, that Pu=a.u and Pv = b.v with a and b distinct and u,v non-zero. Observe that u and v are not in the same direction (linearly dependent) otherwise P would act by the same eigenvalue on each. We also have P(u+v) = c.(u+v) for some c. But P(u+v) = Pu + Pv = au + bv. So au+bv = cu+cv, so (a-c)u = (c-b)v. Since a and b are unequal, this is not a trivial equation (ie it is not 0=0) so u and v are linearly dependent, a contradiction. Hence no such u, v can be found and P has the same eigenvalue, r say, everywhere: that is, P =r.I