# Prove that every operator on an odd-dimensional real vector space has an eigenvalue using induction.

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.

Because if there's an invariant subspace of dimension $$1$$, any nonzero element of that subspace is an eigenvector and we've assumed $$T$$ doesn't have any eigenvectors.