# [Bf-committers] Scale from Matrix

Ramon Carlos Ruiz ramoncarlosruiz at gmail.com
Sat May 6 23:19:30 CEST 2006

```Hi, this is very simple. For Sx first normailze
mat[0][0],mat[0][1],mat[0][2] and create n[0][0],n[0][1],n[0][2], that
contain the rotation only. now pick the bigger number of the first row . .
lets said mat[0][0], then divide that element with n[0][0] and that's Sx or
1/Sx

Un Saludo
RCRuiz

On 5/6/06, Chris Want <cwant at ualberta.ca> wrote:
>
>
> Hrmm, this is quite wrong, please disregard
> (serves me right to do math on a Friday night).
>
> Chris
>
>
> Chris Want wrote:
>
> > Hi Campbell,
> >
> > My linear algebra is a bit shaky, but I think it goes something
> > like this:
> >
> > Suppose you have mat (the one you are feeding to Mat4ToSize),
> > and the positive output from Mat4ToSize, sx, sy, sz.
> >
> > The real SizeX, SizeY and SizeZ should be eigenvalues of mat.
> > The eigenvectors should be the columns of mat (so first column of
> > mat cooresponds to eigenvalue SizeX, etc). Call these column vectors
> > v1, v2, v3.
> >
> > So, what I believe you should do is test that mat*v1 == sx*v1
> > (both sides are column vectors, left side is matrix multiplication,
> > right side is scalar multiplication).
> >
> > If this is true (within some error tolerance), then sx is indeed
> > the correct eigenvalue, so you can set SizeX = sx,
> > else SizeX = -sx (might be prudent to test that mat*v1 == -sx*v1).
> >
> > Similarly, test mat*v2 = sy*v2 and mat*v3 = sz*v3.
> >
> > Anyways, I hope my recollection isn't too hazy, and that the description
> > isn't too incoherent.
> >
> > Regards,
> > Chris
> >
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> >
>
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