Assuming $V$ is finite dimensional and for some invertible $f\in End(V)$ we have a subspace $U$ which is $f$-invariant. Is $U$ also $f^{-1}$-invariant? My reasoning goes as follows:
Obviously $U\subseteq f^{-1}(U)$. If $dim(U)<dim( f^{-1}(U))$ then $\exists w\in f^{-1}(U)\ w\neq0\ \ S.T\ ( f^{-1}(w))=0$. But $ f^{-1}$ is invertible and therefore injective.
But I have not seen this result anywhere else, is this proof valid or am I missing something?
If $f$ is invertible and $U$ is finite dimensional, then $f(U) \subset U$ implies $f(U)=U$ since both sides have the same dimension. Applying $f^{-1}$ yields $f^{-1}(U)=U$. Hence the answer is affirmative.