Re: [voxtech] linear algebra: equivalent matrices
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Re: [voxtech] linear algebra: equivalent matrices
On Wed 07 Dec 05, 10:26 AM, Jonathan Stickel <jjstickel@sbcglobal.net> said:
> Hi Pete; nice to see you on the list again.
Thanks Jon. I finished my degree. Now that I have the official UC
sheepskin that attests to my expertise in physics research, the proverbial
hair can be let down (although now I need to find a job. *ulp*).
> Peter Jay Salzman wrote:
> >Posted to voxtech since this is a CS topic. I'd like to verify some
> >things
> >which I think are true.
> >
> >
> >Consider the set of all square matrices of rank n. The determinant of M,
> >det(M), forms an equivalence class on that set. The equivalence relation
> >is
> >defined by:
> >
> > A ~ B iff det(A) == det(B) (1)
> >
> >
> >Now, like vectors, matrices are always expressed in a basis, whether we
> >explicitly say so or not. So when we write the components M, we should
> >really write M_b where b represents the basis we chose to express M in. We
> >can express M_b in a different basis, say M_a, by a rotation operation:
> >
> > M_a = S^{1} M_b S
> >
> >where S is an orthogonal "rotation matrix". However, no matter what basis
> >we express M in, det(M) remains constant. Therefore, we get an equivalence
> >class on the set of square matrices of rank n based on whether we can
> >rotate
> >one matrix into another. The equivalence relation is defined by:
> >
> > A ~ B iff A = S^{1} M_b S (2)
>
> Did you write this equation correctly? I would expect something like
>
> M_a ~ M_b iff M_a = S^{1} M_b S
Ahh, yeah. You're right. Goofup.
> >for _some_ orthogonal matrix S, which determines the basis for M. There is
> >one rotation matrix S that will make M_b diagonal. That rotation matrix is
> >formed by the eigenvectors of M_b.
> >
> >
> >
> >Big finale:
> >
> >The equivalence classes defined by relation (1) are epimorphic to the
> >equivalence classes defined by relation (2). If we place a restriction on
> >S
> >that it must have a determinant of +1 ("proper" rotations), then the two
> >sets of equivalence classes are isomorphic.
> >
> >What this is really saying is that, when viewed as the sides of a
> >parallelopiped, a matrix will always have the same area no matter what
> >basis
> >you choose to express it in.
> >
> >
> >How accurate is all this? In interested in the lingo as well as the ideas.
>
> Looks fine to me. In my PhD work (almost done!) I deal with 2nd order
> tensors for continuum mechanics theory. I've actually had to specify
> and use rotation matrices explicitly in some of my numerical methods.
>
> Jonathan
I wasn't 100% sure that determinant could also be an equivalent class for
"the same" matrix expressed in different bases. It sounded good, but I
wasn't sure.
Congrats on the degree. It's never done until it's finally done, but I'll
tell you one thing. There is no better feeling than submitting your final
draft. Except for the first time someone actually calls you "dr.
soandso". I got all giggly. ;)
Pete
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