I just developed the following question while thinking about invariance in finite dimensions.
Suppose $V$ is an $n$-dimensional vector space ($2 \leq n < \infty$), defined over some scalar field $F$ that is not necessarily algebraically closed. Suppose $M$ is an $m$-dimensional subspace in $V$, where $0 < m < n$. Suppose $M$ is invariant under some operator $A$ on $V$. Does it follow that $A$ has at least one Eigenvalue?
I see that the case wherein $F$ is algebraically closed is rather straightforward, and it does even require $M$ to be invariant under $A$. That's because, $F$ is algebraically closed implies that $A$ has an upper-triangular matrix w.r.t. some basis in $V$. This means that $A$ has at least one Eigenvalue. I am unsure what happens in the other case when $F$ is not algebraically closed, and would appreciate a pointer. Thanks.
Take $\mathbb{R}^4$. An operator can have the following matrix (in some orthogonal basis): $$ \begin{pmatrix} \cos(\alpha) & - \sin(\alpha) & 0 & 0 \\ \sin(\alpha) & \cos(\alpha & 0 & 0 \\ 0&0& \cos(\theta) & -\sin(\theta) \\ 0&0& \sin(\theta) & \cos(\theta) \end{pmatrix} $$ Take, for example, $\alpha = \theta = \pi/4$. Over $\mathbb{R}$ this operator does not have eigenvalues. But it clearly has two orthogonal invariant subspaces.