Theorem Every operator on an odd-dimensional real vector space has an eigenvalue.
Incomplete proof Suppose that V is a real vector space with odd dimension. The result is obviously true if dim V=1. So now we assume dimension of V is an odd number greater than 1. Assume that the result is true for all real vector spaces with dimension 2 less than dim V. Now we need to prove that T has an eigenvalue. If it does , our work is done. If not, there is a two-dimensional subspace U of V that is invariant under T.
(Doubt) I am aware that every operator on a finite dimensional vector space has an invariant subspace of dimension 1or2. So why not consider two subspaces U and W of dimension 1 and 1 respectively. How we are sure that there will exist a 2 dimensional invariant subspace?
After that proof is very much clear to me.
NB This proof is from "Linear algebra done right'' by Sheldon Axler ,2nd edition chapter 5, last section.