Re: qz

[Stefan van der Walt]

Thanks, Quentin.  I see MATLAB and Octave differs in this regard.

Why is MATLAB's answers numerically so different to that of Octave?
Is the QZ factorisation non-unique?

I am happy to see that the eigenvalues and -vectors are the same,
though.

Regards
Stefan

On Thu, Jul 15, 2004 at 03:02:50PM -0500, Quentin Spencer wrote:

> >octave:5> a = rand(3,3); b = rand(3,3); [AA, BB, Q, Z, V, W] =
> >qz(a,b); Q*a*Z - AA, Q*b*Z - BB
> >ans =
> >
> > -0.506316   0.133307   1.509299
> > -0.425943  -0.168888  -1.856508
> >  0.203352  -0.028521  -0.327870
> >
> >ans =
> >
> > -0.377653  -0.108805   1.424638
> >  0.274852  -0.178300  -0.189498
> >  0.098925   0.051424  -0.448613
> >
> >If I understand correctly, the answers should be much closer to
> >zero(3,3)?
> > 
> >
> You are close. Replace Q with Q', and I get matrices with values on the 
> order of 1E-16.
> 
> -Quentin
> 



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