Suppose that $V$ is a real vector space and $T:V\to V$ is a linear operator having no eigen values.
Prove that every subspace of $V$ invariant under $T$ has even dimension.
Attempt: If $V$ has dimension odd so the characteristic polynomial of $T$ also has odd degree and hence has a real root which is false
Hence $\dim V$= even.
But that's not helping me here,I should deal with invariant subspace here.But how should I do it?