# Prove that any $T$ cyclic-Subspace $U \subsetneq V$ is Eigenspace for diagonalizable $T$

Say we're given a diagonalizable endomorphism $T: V \to V$, $V$ an $n$-dimensional Vectorspace over some Field $F$. Let $U \subsetneq V$ be T-cyclic with generator $u \in U$. I want to prove that $U$ is an eigenspace. (I suspect it is a 1-dimensional Eigenspace).I'm not sure how to proceed. It is easy to show the other direction (any (must it be 1-dimensional? I suspect yes, but not sure) eigenspace is $T$-cyclic), but this direction is harder.

edit: what if I strengthen the hypothesis and say that while $U$ itself is $T$-cyclic, it is not decomposable into proper $T$-cyclic subspaces. Must $U$ then necessarily be an eigenspace (of dimension 1)?

This is not true. Let $V=\mathbb R^3$, and let $T$ be the operator whose matrix in the standard orthonormal basis is given by
$$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix},$$ and let $U=\mathbb{R}^2\times\{0\}$. Then $U$ is a $T$-cyclic subspace with $T$-cyclic vector $$u=\begin{pmatrix} 1\\1\\0 \end{pmatrix},$$ but $U$ is not an eigenspace.
• what if I strengthen the hypothesis and say that while $U$ itself is $T$-cyclic, it is not decomposable into proper $T$-cyclic subspaces. Must $U$ then necessarily be an eigenspace (of dimension 1)? – ghthorpe Jun 2 '17 at 14:36
• I can't say. Perhaps try asking this question in a new post. Note that under this hypothesis, any $u\in U$ is $T$-cyclic. – Aweygan Jun 2 '17 at 18:26