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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

877.0. "A Short Question" by HPSTEK::XIA () Sat May 21 1988 21:13

    Hi there,
         I have a short question.  Hope someone can help me out.  Let
    D = {T: T:R^n --> R^n such that T is linear} (In other word D is
    the set of linear transformation from R^n to R^n--The set of nxn
    matrices).  Now D can be made into a metric space naturally.  Let S 
    be the subset of D such that S consists of all the singular matrices of
    D.  Then S is nowhere dense.  Now since D is a metric space, a measure
    {u,D} can be defined.  Now my question is:  Is u(S) = 0?  My guess
    is that it is indeed true, but can someone give me a proof?
    Thanks in advance.
    Eugene
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877.1any good books on the subject?ZFC::DERAMOI am, therefore I'll think.Mon May 23 1988 03:1553
     This has nothing to do with answering your question.
     The problem itself raised other questions in my mind.
     
>>   Now D can be made into a metric space naturally.
     
     I suppose you just mean that D is "Euclidean n^2 space."
     
>>   Then S is nowhere dense.
     
     I had to look this up to make sure I understood it. 
     The definition means the closure of S does not contain
     a nonempty open set.  How obvious is this here?  Is the
     answer to "is S closed" obvious?
     
     If I understand this right then it is fairly easy using the
     characteristic polynomial of a matrix A (p(x) =
     determinant(A - xI) where I is the identity nxn matrix) to
     show that S is closed and nowhere dense. 
     
     If T is a nxn matrix in D that is in the closure of S,
     then there is a sequence S1, S2, ... of points of S that
     converge to T.  A nxn matrix A is singular if and only
     if zero is a root of its characteristic polynomial. 
     The charactersitic polynomials of the {Sn} converge to
     the charactersitic polynomial of T, and every one has
     zero as a root, and so T has zero as a root:  so T is
     in S, and S is closed.
     
     An open set containing the point A of S also contains
     A - eI for small enough e (e is "epsilon").  This is
     singular only if e is a root of the characteristic
     polynomial of A, but there are at most n such roots,
     so we can find an epsilon that gives a point of the open
     set that is not in [the closure of] S.  So S is nowhere
     dense.
     
     How am I doing so far?  If the above is all garbage please
     let me know.
     
>>   Now since D is a metric space, a measure {u,D} can be defined.     

     There is the Lebesgue measure for D as a Euclidean n^2-space,
     but I didn't realize that there was a way to define a
     measure on an arbitrary metric space.  Could you describe
     the construction?
     
>>   Is u(S) = 0?
     
     Is part of the problem statement: "... for every measure
     {u,D} that can be defined for D"?  Or would a proof for
     Lebesgue measure suffice?
     
     Dan
877.2HPSTEK::XIAMon May 23 1988 19:134
    Re. .1
    I think your proof of S being nowhere dense is fine.  Yes, I meant
    it to be the L-measure when I say naturally.  
    Eugene
877.3one possible approachZFC::DERAMOI am, therefore I'll think.Mon May 23 1988 22:2144
     I once came across the following, given as a "lemma without
     proof": 
     
          If A is a subset of R^2, then A is null if and only
          if for almost all x, A_x is null.
     
     Terminology:
     
          R^2 is the Euclidean plane
          null is "has Lebesgue measure zero"
          almost all is "except on a set of Lebesgue measure zero"
          A_x is {y | (x,y) is in A}
     
     So a subset A of R^2 is null if and only if each slice
     A_x is a null subset of R, except perhaps for a null
     [subset of R] set of counterexample x's.
     
     This should have analogs for other R^n.  So if you consider
     2x2 matrices
     
                        [ a  b ]
                        [ c  d ]
     
     then the set of all 2x2 matrices with a=0 should be null,
     because "A_x" is null for all x except x=0, where here
     "A_x" is the three dimensional slice with a common upper
     left corner.  Call this a 3-1 analog of the lemma.
     
     A 2x2 matrix will be singular if and only if ad - bc = 0.
     This can be broken down into [the first two overlap]:
     
             a = 0 and (b = 0 or c = 0)          [null]
             d = 0 and (b = 0 or c = 0)          [null]
             a ~= 0 and d ~= 0 and ad = bc
     
     Shouldn't the third set be null, because for each point
     (a,d) in R^2 - {(a,d) | ad=0}, the slice A_ad is the
     hyperbola ad = bc in R^2, which is a null subset of R^2?
     [This would be a 2-2 analog of the lemma.]
     
     Does this look convincing for the n=2 case?  How hard
     would it be to generalize to larger n?
     
     Dan 
877.4there he goes again ...ZFC::DERAMOI am, therefore I'll think.Wed May 25 1988 16:548
     Oops, I left part of the lemma out.  It should be
     
        If A is a measurable subset of R^2,
                  ^^^^^^^^^^
     
        then A is null if and only if {x | A_x is not null} is null.
         
     Dan