[voxtech] linear algebra: equivalent matrices
Aaron A. King
aaron.king at umich.edu
Wed Dec 7 12:05:17 PST 2005
Not sure if I'm understanding your question aright, Peter, but I think you're
asking if the equivalence relation (1) is isomorphic to the equivalence
relation (2). That is not the case. If you view a matrix as defining a
parallelopiped in nspace, then the determinant measures the volume thereof.
Now it is easy to have two parallelopipeds which are not congruent but which
have the same volume.
The relation (2) defines a set of equivalence classes which is much finer than
the set defined by (1). To put it another way, the determinant is only one
of many matrix invariants. The full set of invariants under the relation (2)
can be summarized in the Jordan form of the matrix.
Apologies for pontificating, especially if I've answered a question you never
asked.
Cheers (and congrats on the Ph.D.!),
Aaron
On Wednesday 07 December 2005 01:50 pm, Peter Jay Salzman wrote:
 > 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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Aaron A. King, Ph.D.
Ecology & Evolutionary Biology
University of Michigan
GPG Public Key: 0x2B00840F
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