Consider the symmetric bilinear form b on the module M, b : M x M ->
A, where A is a ring and NOT a field. M is possibly a free module. If
B is the matrix with entries b(xi,xj) in A, is B positive
semi-definite? Prove it.

---

Clarification of Question by gcsfred-ga on 07 Jan 2006 18:49 PST

The bilinear form b is non-degenerate. That is, when b(x, y) = 0 for
all y ? X then x = 0 necessarily. For example,

      / 0  1 \
 B' = |      |
      \ 1  0 /

cannot be generated by the bilinear form b.

Even assuming that M is a free A-module, so that the representation of
symmetric bilinear form b by symmetric matrix B (with entries in A)
depends only on the choice of basis (free generators of M over A), one
still lacks in general a definition of positive semi-definite matrix
in this setting.

For example, what would that mean if A were Z/4Z?  The usual criterion:

  u B u' >= 0     (for all vectors u)

requires that A be an ordered ring:

[Ordered ring -- Wikipedia]
http://en.wikipedia.org/wiki/Ordered_ring


regards, mathtalk-ga

The other point to make about this Question, even without knowing more
about the meaning of "positive semi-definite matrix" except that it
should agree with the usual definition in the special case of an
ordered ring (or field), is that the hypotheses stated are
insufficient to establish positive semi-definiteness even in those
narrow cases.

In short the Answer is should be no, B need not be positive
semi-definite, and this can be proven by a simple example.  The
symmetry of M implies symmetry for B, but not positive
semi-definiteness, as may be seen from a bilinear form over ZxZ
represented by the "transpose" matrix:

     / 0  1 \
 B = |      |
     \ 1  0 /

whose eigenvalues are 1 and -1.

regards, mathtalk-ga

gcsfred-ga clarified the Question as follows:

  The bilinear form b is non-degenerate. That is, when b(x, y) = 0 for
  all y ? X then x = 0 necessarily. For example,

        / 0  1 \
   B' = |      |
        \ 1  0 /

  cannot be generated by the bilinear form b.

I'm not sure what X means (capital X) in the above.  Is it the same as
M in the original post?

This example matrix B' represents the symmetric bilinear form:

  b(x,y) = x_1*y_2 + x_2*y_1

over free module ZxZ.  Here b(x,y) = 0 for all y implies x = 0.  I
don't know what more you might want to require about the bilinear form
to try and extract positive semi-definiteness.

Note that by specifying "positive semi-definite matrix" rather than
"positive definite matrix", you suggest a willingness to include
degenerate bilinear forms, such as the one corresponding to:

       / 1  0 \
  B' = |      |
       \ 0  0 /

where b(x,y) = 0 for all y in ZxZ when x = (0,1).

It seems to me that adding non-degeneracy as you've defined it to your
conditions would eliminate the gap between semi-definite and definite,
but do little to advance the proof of positivity.


regards, mathtalk-ga

Thanks for your comments, mathtalk-ga. I'm struggling with the
conventional definition of positive semi-definitiveness for bilinear
forms with rings that are not fields as its codomain for one year and
a half now.
Yes, X in clarification is M in the original question. In adding the
non-degenerate property to the bilinear form, I'm trying to give a
"helping hand" to say its respective matrix is positive semi-definite.
(BTW, I don't understand your notation b(x,y) = x_1*y_2 + x_2*y_1,
since if x ? Z then x is an integer, not a vector. If you are talking
about the domain of the bilinear form, than I understand the notation
(0,1) ? ZxZ, but x = (0,1) makes things confusing because then you are
talking about two "x's": one that ? ZxZ and another that ? Z )

I can't see
       / 1  0 \
  B' = |      |
       \ 0  0 /

(your second B') being generated by a bilinear form that has the
non-degenerancy property. In trying to reconstruct the bilinear form
that formed your second B', I tried b(x,y) = x*y with b(1,1)=1,
b(0,1)=0, b(0,0)=0, b(1,0)=0. The violation to the non-degenerate
property is the pair (1,0) in the domain, because the image is zero,
but x = 1 in this case, whereas it should be zero.

But trying to comment on the problem of the definition of positive
semi-definitiveness, I'm not sure you can find a definition of
positive semi-definitiveness that doesn't assume you are dealing with
a field. I was looking at abstract algebra and category theory, but I
don't quite understand these branches of math.
I think in my research it will be enough to use Milnor's definition of
inner product, which doesn't require the bilinear's codomain to be a
field. ( See http://ww1.math.nus.edu.sg/ProjectArchive/Honours/archive_files/sci01201.pdf
)

Regards,
GF

Please point out any aspect of the notational discussion below that
you think is incorrect or unclear.

For the sake of illustration above M is ZxZ, a free Z-module on two
generators (Z being the usual ring of integers, which happens to be an
ordered ring so that we can define positive definiteness or
semi-definiteness in the usual way).

Bilinear form b (on M) maps pairs of elements in M to "scalars" in Z:

  b: MxM -> Z

An element of M is a pair of integers, e.g. x = (x_1,x_2) or y = (y_1,y_2).

Example of non-degenerate symmetric bilinear form
=================================================

b(x,y) = x_1*y_2 + x_2*y_1

Example of _degenerate_ symmetric bilinear form
===============================================

b(x,y) = x_1*y_1

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

When M is a free module over ring R with finitely many generators,
then M has a direct product representation requiring an equal number
of factors as there are generators.  Thus in the case above we could
treat M as ZxZ.

Generally for module M freely generated by k generators, any symmetric
bilinear form on M (whose elements may be considered k-vectors with
ring entries) can be represented as a symmetric kxk matrix, and this
representation is determined once a choice of basis (generators) for M
over ring R has been fixed.


regards, mathtalk-ga

mathtalk-ga wrote: "[...]which happens to be an
ordered ring so that we can define positive definiteness or
semi-definiteness in the usual way[...]"

Sorry, I don't know this definition of positive (semi-)definite
matrices of ordered ring elements and not field elements. Where can I
find it?

Hi, gcsfred-ga:

You will find the "usual" definition of positive semi-definite matrix
in my first Comment, where I raised the issue of what you meant by the
term in the context of a ring.

Note that this definition has to do with something being "positive"
(or at least nonnnegative), a concept that is not defined for an
arbitrary ring (or even for an arbitrary field).

The real numbers R and the rationals Q are examples of ordered fields.
 The complex numbers are not, nor are any of the finite fields.

So, to go back to the beginning, it really depends on what you mean by
positive semi-definiteness as to whether such a property can be
proven.  I pointed out that in an ordered ring we have the ability to
define positive semi-definiteness because there is a meaningful notion
of when a scalar (aka ring element) is greater than or equal to zero. 
See the link in my first Comment to the Wikipedia page on ordered
rings.  Here's a link about positive definiteness:

[Positive Definite Matrix -- MathWorld]
http://mathworld.wolfram.com/PositiveDefiniteMatrix.html


Without this additional structure, I'm not sure there is any natural
extension of the notion of positive semi-definite matrix.  Of course
you may have another definition in mind that I'm unaware of.  There is
a generalization of positive definiteness to complex Hermitian
matrices, in which complex conjugation is used to generalize the
notion of "symmetric" for real matrices to Hermitian for complex
matrices.


regards, mathtalk-ga

Could that be a gap in the abstract algebra theory that hasn't been
filled? It seems good that for ordered rings you have this notion of
positivity and negativity. That might be sufficient to define positive
definitiveness if you say that the bilinear form is an inner product
that always yields positive scalars, that is, b's image is A+. I just
haven't seen an authoritative source in the literature (NO source as a
matter of fact) with such definition. If I had such definition for
ordered rings then that would be sufficient for me. I would be able to
work with Z for instance. What I can't do is work with fields, in my
particular situation.
Thanks for the comments!
GF