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Generators for Gl(n,R)?

  1. Sep 15, 2011 #1
    Hi, All:
    Just curious as to whether Gl(n,R) (aka invertible nxn matrices over R), is
    finitely-generated by the shear maps : add a k-multiple of one row to another row.
    It seems clear that i) shear maps preserve invertibility (actually preserve determinant),
    and it would seem we could generate any matrix this way. Is this correct?

    Also: are there results for general rings Z, i.e., for a generating set for Gl(n,Z);
    Z any ring?

    Thanks.
     
  2. jcsd
  3. Sep 15, 2011 #2

    Stephen Tashi

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    What you mean by "finitely" generated isn't clear. Are we going to use a finite number of shear matrices?

    Lie groups have "infinitesimal generators", is that the kind of generator you are asking about?
     
  4. Sep 15, 2011 #3
    No, I am thinking more in terms of the generators of a group; like in your first paragraph, a

    finite generating set is one in which a finite set of shear matrices can, under composition,

    produce any invertible matrix. Since the only three operations on matrices that preserve

    invertibility are row-swaps, scaling (by a non-zero factor), and shears, and the

    Jordan decomposition theorem, it seems clear that these three operations (seen as

    matrices ) generate GL(n,R), but, of course, row-swap does not really do anything.

    Still, it seems like we would need infinitely-many shear matrices, since we cannot

    just scale a shear matrix by a constant.
     
  5. Sep 15, 2011 #4

    Stephen Tashi

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    Let's see. Thinking in terms of the matrix representation of GL(n,R), any real number r appears as an entry in some element of Gl(n,R). If we have a finite set of elements S in Gl(n,R) that generates each element of Gl(n,R) then the finite set E of real numbers that are entries in elements of S could be used to generate any real number by some finite process of additions and multiplications.

    In fact, we can consider the non-zero real numbers to be Gl(1,R).) Some expert in the axiomatics of the real numbers can tell us about the propects of generating all real numbers from a finite subset of real numbers.

    A more hopeful question is how well you can approximate the elements in GL(n,R) by a group generated by a finite subset of elements of GL(n,R). That would require defining what we mean by "approximate".
     
  6. Sep 16, 2011 #5

    disregardthat

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    What about GL(n,Z)?
     
  7. Sep 16, 2011 #6

    Stephen Tashi

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    How would that be defined?
     
  8. Sep 16, 2011 #7
    Gl(n,Z) would just be the group of invertible matrices with coefficients in some
    ring Z. A generating set S would then be a subcollection of the set of invertible
    matrices so that every matrix m in Gl(n,Z) is expressible as a product/composition
    of elements in S. In general, I am defining a generating set S for an algebraic structure (A,*)
    to be a subset Ag:={a_si}_i in I ; (for some index I) of A so that every a in A can
    be written as (a_si_1^ei_1)*(a_si_2^ei_2)*...*(a_si_n^e_in) ; ei_j in {-1,1}. e.g., a generating set allows us to obtain every element of the structure by a finite repeated
    application of the operation * on the elements of the generating set. Then a generating
    set for (Z,+), the integers under addition, is {+1,-1}, since any integer n can be
    obtained by a finite sum 1+1+1+1...+1 , or (-1)+(-1)+...+(-1). For a vector space,
    a basis would be a generating set.
     
  9. Sep 17, 2011 #8

    Stephen Tashi

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    It's interesting to contemplate the implied restrictions on the ring in this definition. For example, we can speak of "the set of invertible nxn matrices whose entries are elements of the ring of integers", but the inverses of these matrices won't all have entries that are integers. So the definition only makes sense for special rings.
     
  10. Sep 17, 2011 #9
    Right, I forgot to add the condition that the matrices must be invertible

    over the ring , i.e., a matrix M is in Gl(n,Z) if it is invertible over Gl(n,Z),

    i.e., if it has an inverse with entries in Z. Over the integers these are precisely

    the matrices with determinant +/-1 .
     
  11. Sep 17, 2011 #10

    Stephen Tashi

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    I don't see a simple way to approach the question. It's tempting to start talking about "finitely generated" rings, which I suppose would lead to talking about principal ideals.

    I'm distracted by the entertaining thought of what GL(n,W) would look like if the ring W was GL(n,V), or even if it was GL(n,W) again!
     
  12. Sep 17, 2011 #11
    Well, those are the typesof things done with the Hom functor and in multilinear

    algebra.
     
  13. Sep 20, 2011 #12
    I'm guessing when you are saying "finitely generated" you mean there are matrices M1,M2,M3,...,Mn such that

    M \in [M1,M2,M3,...,Mn]

    for any invertible M? ([M1,M2,...Mn] is the group generated by M1 up to Mn).

    The only other thing I could think you might mean would be

    M=k1M1+k2M2+...+kmMm

    ? (the ki are supposed to be scalars). In this case, you will be able to obtain any matrix you like, although this seems like a scam because you are using the other group operation on these matrices, not the group operation of GL(n,R).

    If you did mean the multiplication one, then I can't see how you would get all of them- all of the matrices will have integer elements and even if you allow yourself to make multiples of the generating elements, you will only get multiples of elements generated by integer matrices.



    In terms of what you mentioned about using the integers instead- there is a group GL(Z,n). You can have a GL(R,n) where R is any unital commutative ring. You just take the obvious definitions and it all works out fine. Nice examples are where you take GL(F,n) where F is a finite field.
     
  14. Sep 23, 2011 #13
    Jamma:

    I think Gl(n,R) -- the entries are real (but their names have been changed, to

    protect the innocent :) ) , is usually taken as a multiplicative group, so a generating

    set would be a collection M:={M_i}i in I , such that every M in Gl(n,R) is the (finite)

    product of elements of M. I thought that transvections (adding a multiple of one

    row to another row), which in R^n are shear maps, would allow us to generate

    every matrix, by reversing Jordan decomposition; in JD , we start with an nxn matrix 'A',

    and, if A is invertible, its Normal form is the identity. This suggests that the set of

    transvections, together with scaling matrices (both used in the decomposition)

    generate Gl(n,R) , but this set is obviously (uncountably-) infinite , ( e.g.,by

    the fact that a scaling can be done for/by any x in the Reals), but the question is

    wether a smaller set (smaller than all transvections and scaling matrices) may do the

    job too.
     
  15. Sep 24, 2011 #14
    Ok, I think I see what you mean. You do have an uncountably infinite set though, and you won't be able to do any better than that, so I'm not quite sure what you mean by a smaller set (although, that's a bit pedantic, you could be looking for a subset of these which still generates GL(n,R)?)
     
  16. Sep 24, 2011 #15
    I wonder if there is a finite set , whether a subset of my proposed generating set
    or otherwise, that will do the job; I wonder if such set exists.
     
  17. Sep 24, 2011 #16
    No, there definitely won't be a finite set because then you can only generate a countably infinite number of matrices in that case.

    Unless you mean you take their multiples too, but I still think that won't give you enough- does the following sound plausible?:
    The finite number of matrices of the generating set on their own will generate some subgroup of GL(n,R) which is countably infinite, so the ratios between the entries is countably infinite too. Allowing multiples only allows all multiples of your matrices, so there will still only be finitely many ratios between the entries if you allow all multiples of your generating set too, but you can construct uncountably many different such ratios for matrices in GL(n,R). So you can't have generated all (or most) or it. Perhaps you can generate a dense set though.
     
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