If exists an Eigenspace $V_{\lambda}$ with dim($V_{\lambda}) \geq 2$, the $f$-invariant subspaces are infinite

Question

Let $V$ a $\mathbb{K}$-vectorial space,with char($\mathbb{K}$) = 0.

Is it true that if exists an Eigenspace $V_{\lambda} \subset V$, with dim($V_{\lambda})\geq 2$, the $f-$invariant subspaces are infinite ?

If so,for which dimension ?

Any help or sketch of the proof would be appreciated.

Answer 1

Per the comment below your question, the question you seem to be asking is

If $f: V \to V$ is a linear map with an eigenspace $V_{\lambda}$ of dimension at least $2$, then $f$ has infinitely many invariant subspaces.

The answer to this question is yes. Note that if $V_{\lambda}$ is an eigenspace, then any subspace $U \subset V_\lambda$ is an eigenspace and thus an $f$-invariant subspace.

If $V_{\lambda}$ has dimension at least $2$, then we can find infinitely many one-dimensional subspaces $U$. One construction of such a family of subspaces is as follows: suppose that $v_1,\dots,v_d$ is a basis of $V_{\lambda}$. For $t \in \Bbb K$, define $$ U_t = \operatorname{span}(\{v_1 + tv_2\}). $$ Note that if $t_1 \neq t_2$, then $U_{t_1} \neq U_{t_2}$.

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We can extend this construction as follows. For any $k < d = \dim(V_{\lambda})$, define $$ U_t = \operatorname{span}(\{v_1 + tv_2,v_3,\dots, v_{k+1}\}). $$