Let $ G $ be a linear algebraic group. Is it true that a subgroup $ H $ of $ G $ is Zariski closed if and only if there exists a representation $ \pi: G \to \mathrm{GL}(V) $ and a vector $ v \in V $ such that the stabilizer $ G_v:=\{g \in G: \pi(g)v=v \} $ is equal to $ H $?

I think one implication is clear since $ G_v $ is certainly a subgroup and the equation $ \pi(g)v=v $ is polynomial in the matrix entries. Thus any stabilizer must be Zariski closed.

I am not sure of the reverse implication. Is it really true that every Zariski closed subgroup of $ G $ arises as the stabilizer of some vector in some representation?