# Gauss decomposition of an algebraic group.

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?

• The set of $n \times n$ matrices with non-vanishing principal minors is not a group. E.g. if $U = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ and $L = \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}$ then $L U L^{-1} =\left(\begin{array}{rr} 0 & 1 \\ -1 & 2 \end{array}\right)$. Dec 16, 2015 at 16:20

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 $G$.
• @Loeffler: I understand from your argument that the statement in general is false. But what should be the assumptions on $G$ to make such a statement true? Dec 16, 2015 at 14:57
• If you take that viewpoint, then the "Gauss decomposition" does not exist even for $\mathrm{GL}_n$ (except when $n = 1$). You'll have some real trouble decomposing $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Dec 16, 2015 at 16:19