Let $G=GL\_n(K)$ where $K$ is an algebraically closed field of characteristic zero. Let $V$ be a finite dimensional rational representation of $V$. 
Assume that $v\in V$ has a reductive stabilizer $H\subseteq G$.
I would like to ask for a reference for the following fact: 



> 
> There is a rational finite dimensional representation $W$ of $
> G$ and a point $w\in W$ such that the stabilizer of $(v,w)$ in $V\oplus W$ is also $H$, and such that the orbit $G\cdot (v,w)$ is closed in $V\oplus W$.
> 
> 
> 


A simple instance of the above statement is the case where $n=1$, $V=K$ is the one dimensional representation upon which $x\in G$ acts by $x$, and the element $v$ is $1\in K$. The representation $W$ will then be $V^\*$, and $w\in W$ can be any nonzero vector. The orbit of $(v,w)$ is then the hyperbola in $K^2$ which is closed (even though the orbit of $v$ is not closed).


Is it known if this is also true in case we replace $GL\_n$ with some other reductive group?



Closed orbits for the action of general linear groups
I don't know a reference to the question as asked, but below I give a brief argument.


Let $G$ be an algebraic group and $H$ a reductive subgroup.
By Matsushima's theorem, $G/H$ is affine. Theorem 1.12 in Borel's "Linear algebraic groups" tells that there any $G$ affine action could be $G$-equivariantly closedly embedded in a rational $G$ representation.
Applying to $G/H$ we obtain a $G$-representation $W$ and $w\in W$ having stabilizer $H$ and a closed orbit isomorphic to $G/H$.


Given any $G$-representation $V$ and $v\in V$ with stabilizer $H$, the stabilizer of $(v,w)\in V\oplus W$ will be $H$. We are left to argue that the orbit will be closed. Since $Gw\subset W$ is closed, it is enough to show that $G(v,w)\subset V\times Gw \simeq V\times G/H$ is closed. For this you need that the $H$-orbit of $v$ is closed, which is clear.


