Let $K$ be any field. Consider $GL_n(\mathbb{K})$ as an algebraic group. I know that it has a Gauss decomposition, i.e $GL_n(K)=I^- D I^+$, where $I^-$ and $I^+$ are the lower unipotent matrix and the upper unipotent matrices respectively and $D$ is the set of diagonal matrices in $GL_n(K)$. My question is the following:
Is the following statement true?
For any algebraic subgroup $G$ of $GL_n(K)$ we have the decomposition
$G=(I^-\cap G) (D \cap G) ( I^+ \cap G)$
I know that this is true when for example $G=SL_n(K)$.
Thanks in advance for help and reference.
Added: I think for $SL_n$ or $GL_n$ the above decomposition are incorrect. In order to get the decomposition one should work with their subgroups consisting of matrices with non-trivial principal minors.
After the answer by David my question is:
For what conditions on $G \subset GL_n(K)$ the gauss decomposition is possible?
This is false. Let $K = \mathbf{R}$, $n = 2$ and let $G$ be the matrices of the form $\begin{pmatrix} x & y \\ -y & x \end{pmatrix}$ with $x^2 + y^2 = 1$. Then $I^- \cap G = I^+ \cap G = \{1\}$ and $D \cap G = \{ \pm 1\}$, so the product $(I^- \cap G) (D \cap G) (I^+ \cap G) = \{\pm 1\}$ is much smaller than $G$.