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Singular Matrices

  1. Mar 9, 2004 #1
    I am getting all singular matrices in my research on the quantization of one-dimensional space.

    I think this is the proper place for me to ask all mathematicians the disadvantage of singular matrix.

    I know that singular matrix does not have an inverse since its determinant is zero and hence a metric cannot be defined.

    What is the parctical use of a metric besides giving a "distance?"

    Zero metric can also mean zero mass in physics. But if all masses are zeros, where do the experimentally determined masses come from?
     
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  3. Mar 9, 2004 #2

    matt grime

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    Matrices do have a metric (many in fact) irrespective of whether or not they are singular, the only time d(M,N) is zero if d is a matrix metric is if M=N


    metrics yield topologies, but as a metric is a (generalized) notion of 'distance', I'm not sure I understand what you're asking.

    singular of a matrix means that the codimension of the image is not zero, that do you for another interpretation? or the dimension of the kernel is non-zero.

    anyway, just becuase the determinant is not a metric does not imply there is no metric in the space of matrices. There is an impotant one, and it implies that if M is an invertible matrix, and H is a matrix such that d(M,M+H) < 1/d(M,0) that M+H is invertible too.
     
  4. Mar 9, 2004 #3
    matt grime,

    Thanks for helping me again with my math deficit.

    Are the following square symmetric matrices singular and do they have metric, inverse, etc.?

    [tex]H^{+} = \left(\begin{array}{cc}+1 & -1\\-1 & +1\end{array}\right)[/tex]

    and

    [tex]H^{-} = \left(\begin{array}{cc}-1 & +1\\+1 & -1\end{array}\right)[/tex]

    Isn't nonzero determinant a necessary condition for having a metric?

    Please clarify further?

    Antonio
     
  5. Mar 9, 2004 #4

    matt grime

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    both those are singular matrices yes, but there are still metrics on (2x2) matrices and the metric is not zero on them.

    let [a,b,c,d] denote the matrix

    |a b|
    |c d|

    then one norm is

    n([a,b,c,d]) = sqrt(|a|^2 + |b|^2 + |c|^2 + |b|^2)

    another is the usual operator norm:

    n(M) = sup ||M(x)||

    where sup is taken over all vectors x of norm 1.

    A metric can be defined from a norm:

    d(A,B) = n(A-B)


    neither matrix has an inverse, but that isn't important really.

    nxm matrices are just nxm dimensional real space, where there are norms too.


    det isn't a metric - det can be -ve, for instance.

    perhaps you should say what you think a metric is?
     
  6. Mar 9, 2004 #5
    matt grime,

    Thanks for your continued elucidations.

    I am still having difficulty understanding your math notations and their meanings.

    The truth is I don't know what a metric is. At first I thought metric defines a "distance" in some space.

    This might be a stupid question, can there be a metric in one-dimensional space, in two-dim space, in 3-dim space?

    The important question for me is how do we define a metric in one-dim space using a matrix?

    Antonio
     
  7. Mar 9, 2004 #6

    selfAdjoint

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    If X is some space, a metric is a map [tex]d: X \times X \rightarrow R [/tex] that associates each pair of points in X to a real number, that satisfies

    [tex] d(x,x) = 0 [/tex]

    [tex]d(x,y) = d(y,x)[/tex]; and

    [tex]d(x,z) \le d(x,y) + d(y,z) [/tex]

    As long as they obey these laws, metrics don't have to be nice, and there are metric spaces that don't have anything to do with euclid.
     
    Last edited: Mar 9, 2004
  8. Mar 9, 2004 #7
    selfAdjoint,

    Your notations are a lot better than matt's but still I'm having troubles with your math logic.

    Is d:X x X -->R a one-dim metric? What is the operator x? Is it multiplication?

    How to write the map for a 2-dim metric, or a 3-dim metric?

    There is a metric in GR where the g11 is 1, g22=-1, g33=-1, g44=-1 and all the other elements are zeros. What is the meaning or purpose of this metric in GR? (another stupid question).

    Antonio
     
  9. Mar 9, 2004 #8

    matt grime

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    the small x there means cartesian product.

    simply what is being got at is that the distance notion requires two inputs - the initial point and the final point; the dsitance is from a point to another point.

    I'm sorry that you find my notations unclear, but they are very standard.


    you ought to know what a norm is and what sup means; they are easily learned ideas; wolfram can help, usually.


    a metric satisfies 3 rules:

    two points are zero distance apart iff they are the same point

    the distance from a to b is the same as the distance from b to a,

    the distance from a to b directly is shorter than going via nay intermediate point.



    How does one define a metric in 1d using a matirx? dunno, because the only matrices from R to R are 1x1, ie R again, and the only singular 1x1 matix is 0... and i've no idea what you are getting at.

    i though you wanted a metric on some rxp matrices. of which there are infintely many.

    what do you mean by an n-dim metric? metrics do not have a dimension.


    i really don't understand what's wrong with my notation; in fact i'm more than slightly offended by the implication. perhaps you should learn what the meanings of the terms you presumne to use are first?
     
    Last edited: Mar 10, 2004
  10. Mar 10, 2004 #9

    selfAdjoint

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    [tex]g^{\mu \nu}[/tex] is the metric tensor. The metric between two points [tex]u_{\mu} [/tex] and [tex] v_{\nu}[/tex] is [tex]g^{\mu \nu}u_{\mu}v_{\nu}[/tex] where the Einstein summation condition is observed, i.e. you add up all the terms for [tex]\mu, \nu = 0, 1, 2, 3 [/tex]. This gives you a unique real number. You can verify for your self with examples that this definition of a metric - a real number associated with each pair of points - satisfies the three metric axioms I stated.
     
    Last edited: Mar 10, 2004
  11. Mar 10, 2004 #10
    matt and selfAdjoint,

    Thanks to both of you for the valuable helps. These will help me save time but still I have to go on WolframResearch - Eric Weisstein's world of Math and Physics.

    At least now I know what to look for. Thanks again.

    Antonio
     
  12. Mar 10, 2004 #11
    matt grime,

    you quote
    _________________

    i really don't understand what's wrong with my notation; in fact i'm more than slightly offended by the implication. perhaps you should learn what the meanings of the terms you presumne to use are first?
    __________________

    There is nothing wrong with your notations, I just can't understand them. The faults is mine. I am trying to learn the meanings and the notations simultaneously and to me it does present a problem. If I stumble on Egyptians hieroglyphics, I would have the same problem. I know there is nothing wrong with the hieroglyphics, the message is there but it is in codes.

    Antonio
     
  13. Mar 10, 2004 #12

    matt grime

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    ok,my reply was far too nasty in tone anyway and i was going to take it down, or at least edit it.
     
  14. Mar 10, 2004 #13
    matt grime,

    It's too late for that, the whole world (the entire universe) saw it.

    Antonio
     
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