I'm trying to understand a proof from my textbook. They are showing that every operator on a finite dimensional real vector space with odd dimension has an eigenvalue. I already know that every operator on a finite dimensional real vector space has an invariant subspace of dimension two or one...
We have , which is the finite dimensional vector space, and , which is an operator on V. There is a subspace of that is invariant and has dimension 2 (otherwise we are done). There is some other subspace such that .
Anway, here is what I don't understand. Don't we know that either is invariant under or is non-injective (and has eigenvalue 0?). If and , then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a such that , which shows that isn't injective?
The short version: if W is not invariant under T, then must T have eigenvalue 0?