A matrix group and quotient spaces This is a 2 part question.


*

*I have been studying a particular matrix group $G \le GL(n,\mathbb R)$ with $n \ge 3$ and I managed to show elements of my group $A \in G$ have the block structure
$$
A = \left(
\begin{array}{cc}
O(3) & 0 \\
A_{21} & A_{22}
\end{array}
\right)
$$
Now $A_{22}$ must be invertible since I started with $GL(n, \mathbb R)$. So is it true that this matrix is an element of the group $G = O(3) \times GL(n-3,\mathbb R)$? I'm not sure how to account for the "extra" elements $A_{21}$ when writing $G$ as a direct product of groups.

*I have a function
$$
f:\mathbb R^n \rightarrow \mathbb R
$$
which happens to be invariant under the action of my group $G$. In other words, $f(Ax) = f(x)$ for each $A \in G$. There is a comment in a thread here which says if I want to minimize my function, I only need to search for solutions in the quotient space $\mathbb R^n / G$. I am having trouble understanding what this quotient space looks like - can anyone provide some intuition how to go about visualizing this quotient space?
 A: (1) We may write the matrix group as a semidirect product
$$(\textrm{O}(3) \times \textrm{GL}(3, \Bbb R)) \leftthreetimes \textrm{M}(n - 3, \Bbb R) .$$
Explicitly, the homomorphism $\phi : \textrm{O}(3) \times \textrm{GL}(n - 3, \Bbb R) \to \textrm{Aut}(\textrm{M}(n \times 3, \Bbb R))$ defining this product is $\phi(A, B)(C) = B C A^{-1}$.
(2) I'll assume you're working with the action given by restricting the standard action of $\textrm{GL}(n, \Bbb R)$ on $\Bbb R^n$ to $G$. It is useful to write this action in a block decomposition, as
$$\pmatrix{A & 0 \\ C & B} \cdot \pmatrix{{\bf x} \\ {\bf y}} = \pmatrix{A {\bf x} \\ C {\bf x} + B {\bf y}}.$$
Since $\textrm{O}(3)$ acts transitively on $\Bbb R^3$, we may use an element of the form $\pmatrix{A & 0 \\ 0 & I_{n - 3}} \in G$ to map any element $({\bf x}, {\bf y}) \in \Bbb R^n$ to the element $((\lambda, 0, 0), {\bf y})$, $\lambda := |{\bf x}|$. 
Similarly, since the orbits of $\textrm{GL}(n - 3, \Bbb R)$ on $\Bbb R^{n - 3}$ are the zero singleton and its complement, we may use an element of the form $\pmatrix{I_3 & 0 \\ 0 & B}$ to then bring an element of $\Bbb R^n$ to one of the form $((\lambda, 0, 0), (\epsilon, 0, \ldots, 0))$, where $\epsilon \in \{0, 1\}$.
If $\lambda \neq 0$, the element with $A = I_3$, $B = I_{n - 3}$ and $C = \lambda^{-1} E_{11}$ (where, $E_{11}$ is the $(n - 3) \times 3$ matrix with $(1, 1)$ entry $1$ and all other entries $0$) maps $((\lambda, 0, 0), (\epsilon, 0, \ldots, 0))$ to $((\lambda, 0, 0), {\bf 0})$. Thus, any $G$-orbit contains an element of one of the following forms:
$$((\lambda, 0, 0), {\bf 0}), \quad \lambda \geq 0 \textrm{;} \qquad ({\bf 0}, (1, 0, \ldots, 0)) .$$
On the other hand the $G$-action preserves the length of the projection of an element of $\Bbb R^n$ onto its first three components, and the $G$-orbit of $0$ is $\{0\}$, so the above list parameterizes the $G$-orbits, that is, no two elements of the list are in the same $G$-orbit.
A: For the first part, the answer is no. The direct product is the group
of block matrices
$$\pmatrix{A_{11}&0\\0&A_{22}}$$
with $A_{11}\in O(3)$ and $A_{22}\in GL(n-3,\Bbb R)$.
