Let $G$ be a reductive group over an algebrically closed field $\mathbb{K}$ acting on a finite dimensional vector space $V$. Let $\lambda$ be a one-parameter subgroup of $G$. Let's consider for $r \in \mathbb{N}$ $$ V_r = \{ v \in V \, | \, \forall t \in \mathbb{K}^*, \, \lambda(t) \cdot v = t^r v \}. $$ Let's denote $GV_r = \{ g \cdot v \, | \, g \in G, v \in V_r \}$.
I wanted to show that $GV_r$ is closed. But I think that it is not a straighforward result. Since even if $V_r$ is closed, that doesn't imply that $GV_{r}$ is closed. For example, we know that an orbit of an element is locally closed but not necessarily closed.
In addition, I think that the fact that $G$ is reductive play a role in the proof, but I don't know how.
Since, I'm a begginer in the field of Algebraic groups, I'd really want some help to prove that $GV_r$ is closed. Any hint or reference would be great.
Thanks in advance for your enlightenment.
K. Y.