generalisation of GL(3,R) polar decomposition Does polar decomposition work when the 'orthogonal' matrices are not orthogonal wrt to the identity ($O^TO=Id$), but wrt to some other symmetric matrix $K$ (i.e. $O^TKO=K$)?
Specifically, $GL(3,R)$ can be decomposed as $SO(3)\times Diag(3)\times SO(3)$, but can it also be decomposed as $SO(3)\times Diag(3)\times SO_K(3)$? (SO_K are the K-orthogonal matrices mentioned above)
Thanks for any help.

EDIT:
Sorry to resurrect this, but I've had a further thought. First of all thank you for you answers (over my head as they were, they're still appreciated). From what I gather the $SO_k \times Diag^+ \times SO_k$ decomposition works, but $SO \times Diag^+ \times SO_k$ does not? 
What if we were to use the standard polar decomposition on $Q\in GL(3)$ so that $Q=RS$ for orthogonal $R$ and symmetric $S$, and then further decompose this symmetric matrix into a  $SO_k \times Diag^+ \times SO_k$ product? This should be valid, since $S\in GL(3)$, right? So you'd end up with a matrix that looks like $SO\times [something]\times SO_k$. This is really what I was after when I first asked the question (I realise I could have been clearer about that). The something in the middle doesn't really matter to me.
Does that make any sense?
Thank you again.
 A: Let $G = SL_3(\mathbb{R})$.  Let $K = SO_3(\mathbb{R})$ and $H=SO_{2,1}(\mathbb{R})$ be the two orthogonal groups in three variables (one can think of $H$ as the image of $SL_2(\mathbb{R})$ under the symmetric square or adjoint representation).  Let $A$ be the subgroup of diagonal matrices.  Let $g\in G$ be arbitrary.
I think the problem is then whether either of $G = K^gAK$ and $G=H^gAK$ might hold.
The first assertion is false: I will show that $G = K^g AK$ iff $g\in AK$.  The sufficiency is clear.  For necessity let $S=G/K$ be the symmetric space with $x_0$ the point fixed by $K$.  Then $F_0 = A\cdot x_0$ is the standard flat through $x_0$, and $K^g A K$ is the preimage in $G$ of the union of the set $K^g F_0$ of flats. Let $x_g = gx_0$ be the point fixed by $K^g = gKg^{-1}$.  Then $g\notin AK$ is equivalent to $x_g \notin F_0$.  Since $F_0$ is closed there is a ball $B_S(x_g,r)$ which is disjoint to $F_0$. This ball is $K^g$-invariant, so it is disjoint from the union of the $K^g$-translates of $F_0$ as well.
I'm not sure about the second assertion.
