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.