it might have 3 real solutions. but say it had an invariant 2-dimensional eigenspace. then certainly that eigenspace has a one-dimensional invariant subspace (just pick one of the eigenvectors). or more, simply, if v is an eigenvector of f corresponding to any real eigenvalue (and there must be at least one) λ, then:
f(cv) = c(f(v)) (since f is linear)
= c(λv) = (cλ)v. so if U = {av: a in R}, then f(U) ⊆ U, hence U (which is a line going through the origin in the direction of v) is invariant under f.