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?

Thanks.