Re: [vox-tech] linear algebra: equivalent matrices

[Jonathan Stickel <jjstickel@MAPSsbcglobal.net>]

Hi Pete; nice to see you on the list again.

Peter Jay Salzman wrote:
> Posted to vox-tech 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

> 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


> Thx!
> Pete
>
>
> PS- Whether a rotation is S^{-1} M_b S or S M_b S^{-1} depends on how your
> favorite linear algebra author defines his/her rotation matrices.
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