Let $G$ be a linear algebraic group (a closed subset of $GL(n,K)$, for some algebraically closed field $K$) and define $\dim(G)$ as its dimension as an algebraic variety over $K$.
Is it true that this concept of dimension is independent of the choice of a basis? In other words, is it true that $\dim(G)=\dim(x^{-1}Gx)$ for every $x\in GL(n,K)$?
I think this should be true but I've found no reference to this fact in any basic text on the subject (but maybe I've missed some nice texts). It seems a natural question to me but I fail to see how change of basis is connected with the dimension as an algebraic variety.
The isomorphism is $\Phi\colon G\subseteq GL(n,K)\to x^{-1}Gx\subseteq GL(n,K)$ sending $z\mapsto x^{-1}zx$.
Of course this map (and it's inverse) is a morphism of algebraic because the multiplication on $GL(n,K)$ is a morphism.