# 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 an odd dimension.

The result is obviously true if $$\dim V=1$$. So now we assume the dimension of $$V$$ is an odd number greater than $$1$$. Assume that the result is true for all real vector spaces with dimension equaling $$\dim V-2$$.

Now we need to prove that $$T$$ has an eigenvalue. If it does, our work is done. If not, there is a $$"-D$$ subspace $$U\leqslant V$$ that is invariant under $$T$$.

(Doubt) I am aware that every operator on a finite-dimensional vector space has an invariant subspace of dimension $$1$$ or $$2$$. So why not consider two subspaces $$U$$ and $$W$$ of dimension $$1$$ and $$1$$ respectively? How are we sure that there will exist a $$2$$-dimensional invariant subspace?

After that proof is very much clear to me.

NB

Source: Alexander, Sheldon, 'Linear algebra done right', $$2^{\text{nd}}$$ 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.