[vox-tech] linear algebra: equivalent matrices

[Peter Jay Salzman]

On Wed 07 Dec 05,  3:05 PM, Aaron A. King <aaron.king at umich.edu> said:
> 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 n-space, 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

Don't apologize -- this is exactly what I wanted to know!

OK, so then it's not true that all matrices with the same determinant are
related by a rotation.  I was going to ask about trace too,

   Tr[M_b]

      = Tr[S^{-1} M_a S ]

      = Tr[S] Tr[M_a^{-1}] Tr[S^{-1}]

      = Tr[M_a^{-1}]

      != Tr[M_a]

but trace isn't invariant under inverse, so the trace relation can't be
equivalent to relation (2) either.

Thanks for chiming in!
Pete








> On Wednesday 07 December 2005 01:50 pm, Peter Jay Salzman wrote:
> | > Peter Jay Salzman wrote:
> | > >
> | > > 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} B S                (2)
> | > >
> | > > 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.