Let V be a vector space (over $\mathbb{F}=\mathbb{R}$ of $\mathbb{C}$)
of dimension $n\geqslant 2$, such that $f$ has only
$\{0_V\}$ and $V$ as it's invariant subspaces. Examine if $f$ has eigenvalues.
Attempt. If $\lambda\in \mathbb{F}$ is an eigenvalue of $f$, then eigenspace
$V_f(\lambda)\neq \{0_V\}$ is invariant subspace of $f$ and $V_f(\lambda)=V$, by hypothesis, i.e. $f=\lambda I_V$ ($I_V(v)=v$ the identity map over $V$). So we would say that either $f$ has no eigenvalues, either $f=\lambda I_V$ and $\lambda$ is it's unique eigenvalue.
Is the argument correct?
Thanks in advance.
Best Answer
An eigenvector of $f$ can be defined as a vector spanning a $1$-dimensional $f$-invariant subspace. Since your hypotheses exclude such subspaces, there can be no eigenvectors (and hence no eigenvalues).