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What what does it mean for the set of invertible nxn matrices to be open?

  1. Mar 4, 2009 #1
    Real Analysis: What does it mean for the set of invertible nxn matrices to be open?

    1. The problem statement, all variables and given/known data

    The set of invertible nxn matrices is open in M, is it also dense?


    2. The attempt at a solution

    Let S = set of invertible nxn matrics, S contained in M

    Let A be any invertible element in S. We want to show that there exists an open ball of ivertible elements centered at A. Let 0 < d < det^-1(A)

    But I have ho idea what it means for the set of invertible nxn matrices to be open or how to proceed from here. Any help would be great!
     
    Last edited: Mar 5, 2009
  2. jcsd
  3. Mar 5, 2009 #2

    Mark44

    Staff: Mentor

    To talk about a set being open in M (a metric space -- you didn't say), there has to be a way to measure distance. How is the distance d between two matrices defined in M? It can't be the determinant, because that's defined only on a single matrix.

    Can you give us the full explanation of this problem? The way you have stated it, it doesn't make any sense to me.
     
  4. Mar 5, 2009 #3

    Office_Shredder

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    Gold Member

    The standard way of setting a norm on matrices is by taking the square root of the sum of the square of its entries. Essentially, you treat the set of nxn matrices as the set Rn^2.
     
  5. Mar 5, 2009 #4

    Dick

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    There's a number of ways of defining a matrix norm. The important property they have is that the determinant function from nxn matrices to R is continuous.
     
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