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.


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.

  • $\begingroup$ psoitive definite symmetric matrix and $SO_K$ on both sides, than I'd say yes. $\endgroup$
    – Marc Palm
    Jul 13, 2011 at 16:27
  • $\begingroup$ @pm, even with those restrictions, you'd need to conjugate the group of diagonal matrices, no? (Also, I'm bothered by the product notation, since this isn't even a product of topological sets. I think that can be fixed by restricting to diagonal matrices with positive determinant, though.) $\endgroup$
    – LSpice
    Jul 13, 2011 at 17:07
  • $\begingroup$ @Spice, I wanted to say, if $K$ induces a scalar product on $\mathbb{R}^3$, the decomposition result is the same, since all scalar products are the same, since there exists $\sqrt{K}$. Then every matrix $m$ in $End(\mathbb{R}^3)$ can be given as $m=o d o'$ for $d$ diagonal and $o,o'\in SO_K(3)$. $\endgroup$
    – Marc Palm
    Jul 13, 2011 at 18:16
  • 1
    $\begingroup$ @pm: what holds for GL(2)? the SO(2)xDiag(2)xSO_K(2) decomp.? No I'm not sure at all. 3 dimensions was the only example I had been thinking of. I should have said K is positive definite, sorry. So basically its either SO(3) on both sides or SO_K(3) on both sides and the mixed decomposition is impossible? $\endgroup$
    – em12
    Jul 13, 2011 at 19:31
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    $\begingroup$ Why do you call this polar decomposition? It looks like singular value decomposition. (See Wikipedia) $\endgroup$
    – S. Carnahan
    Jul 14, 2011 at 2:05

1 Answer 1


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.

  • $\begingroup$ Ooh, an earlier remark I made reflected inattentive reading of the question. There are two different questions being asked, and they are not equivalent. In case anyone who doesn't understand the situation is still listening, the point is that, on one hand, $G=KAK$ for $K$ the orthogonal group of any positive-definite form, by the spectral theorem for (corresponding) hermitian operators. On the other hand, asking for $G=K_1AK_2$ is asking too much, as Lior's answer shows. Perhaps part of the issue is understanding that the two forms of question are not the same. $\endgroup$ Jul 13, 2011 at 23:15
  • $\begingroup$ How do you show that $G=KAK$ if $A$ remains the standard split torus but $K$ is an arbitrary maximal compact? The proofs I know assume that the fixed point of $K$ in the symmetric space lies on the flat associated to $A$, or equivalently that there is a Cartan involution which fixes $K$ and is the inverse on $A$, or equivalently that $G=NAK$ is an Iwasawa decomposition. $\endgroup$ Jul 14, 2011 at 21:40
  • $\begingroup$ @Lior, my geometric picture here is unfortunately quite weak, but I have the feeling that the question you ask is a much more sophisticated version of my comment above (where I wondered whether we needed to conjugate the split torus when changing the form). $\endgroup$
    – LSpice
    Jul 15, 2011 at 14:48
  • $\begingroup$ @paul: I think I'm getting it -- if $K,A$ are the standard subgroups then $G=KA′K$ where $A′$ is any conjugate of $A$. I found a proof for $GL_2(\mathbb{R})$: by Iwasawa it's enough to consider $A′=nAn^{−1}$ for some $n\in N$. Then the $A′$-orbit of $i$ in the upper half-plane is a diagonal line $y=1+mx$ ($z=i+(1+mi)x$) and rotating it about $i$ will sweep the whole plane. How does the general argument work? $\endgroup$ Jul 16, 2011 at 16:34

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