## Transformation(s) that “fixes” vector

I have defined the stretching of a vector as y=vx where v = stretch (in GL+(3,R)) & x, y = vectors. However, someone kindly has pointed out that this doesn't define v uniquely as vu where u = transformation fixing x (i.e. ux = x) results in same stretch since (vu)x = v(ux) = vx. The suggested solution is to use "the representative of v in the quotient of GL(3) by the subgroup of GL(3) that fixes x." I'm wondering if that subgroup generally is known. If not, could I just use equivalence classes of v ?

Thanks.

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 Yes - it's isomorphic to $GL(2)$ and acts on the plane to which $x$ is normal. – David Roberts Mar 5 2011 at 0:02