Invariant hyperplanes that does not pass through the origin Let $V$ be a vector space. Let $\alpha$ be an endomorphism of $V$. Under what circumstances does there exist a proper subspace $E<V$ and $\mathbf d\in V\backslash E$ such that the hyperplane $E+\mathbf d$ is invariant under $\alpha$?
I know that for some $2\times 2$ matrices, the invariant line does not pass through the origin. In general, if $1$ is an eigenvalue of $\alpha$, then there exists an invariant hyperplane which does not include $0$. However, having $1$ as one of the eigenvalues seem to be just a sufficient condition. Can anyone give a necessary and sufficient condition?
I am willing to limit the discussion to $V=\mathbb R^n$ or $\mathbb C^n$ if necessary.
 A: Assume a linear transformation, $\alpha:V\to V$ and $d\neq 0$ and subspace $E$ such that $E+d$ is invariant.
We claim that $E$ must be invariant.
Indeed, if $\alpha(E+d)=E+d,$ then $d-\alpha(d)\in E\cap \alpha(E)$.
Therefore, if $v\in E\setminus T(E),$ then $v+d= v'+T(d)$ for some appropriate $v'\in T(E),$ but then $v=v'+T(d)-d\in T(E)$ by assumption, which is a contradiction. The other inclusion is similar, or you could simply say that $dim(E)\geq dim(T(E))$, but we just showed $E\subseteq T(E)$.
Now, we know that $\alpha(d)=d+w$ for some $w\in E$, so a necessary condition is the existence of an invariant subspace $E$ and a linearly independent one-dimensional subspace $E'$such that $\alpha(E')\subseteq E\oplus E'$.
In case $V$ is finite dimensional over $\mathbb{C},$ apply the Jordan Normal Form to $\alpha$ restricted to $E\oplus E',$ yielding a basis of generalised eigenvectors. Note that one such generalised eigenvector $d'$must lie outside $E$ and hence $d'=\lambda d+w'$ for some $\lambda\in \mathbb{C}$ and $w'\in E$.
Then, $\alpha(d')=\lambda d+w+\alpha(w'),$ and $w+\alpha(w')\in E,$ implying that, in fact, $d'$ must be a generalised eigenvector with corresponding eigenvalue $1$.
As you seem to know that this is sufficient (which should also follow from the above), $\alpha$ has these properties if and only if one of its eigenvalues is $1$.
