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Space of matrices with non-zero determinant

  1. May 13, 2014 #1
    Hi there!
    How can I prove that the space of matrices (2x2) nonzero determinant is dense in the space of matrices (2x2) ?

    I've already proved that it's an open set.
    Thanks.

    PD: Sorry about the mistake in the title.
     
    Last edited: May 13, 2014
  2. jcsd
  3. May 13, 2014 #2
    The space you named isn't dense or open. In fact, it's closed with empty interior, i.e. its complement is open and dense.

    [For any people reading, the named space has since changed. Now it is open and dense.]
     
    Last edited: May 13, 2014
  4. May 13, 2014 #3
    Sorry, you're right it's not the matrices with determinant = 0, I should say nonzero determinant.

    Thanks!!
     
    Last edited: May 13, 2014
  5. May 13, 2014 #4
    There are lots of different ways you could show it. One would be to consider, for any matrix ##M## and small nonzero number ##\epsilon## the nearby matrix ##M_\epsilon = M + \epsilon I##. What is the determinant of ##M_\epsilon##? Can you show it's nonzero for small but nonzero ##\epsilon##?
     
  6. May 13, 2014 #5
    The determinant of [itex]M_\epsilon[/itex] is always nonzero. Unless [itex]M = \begin{pmatrix} -\epsilon & 0\\ 0 & -\epsilon \end{pmatrix}[/itex].
    Is [itex]\epsilon[/itex] fixed or just a number decreasing to zero?
     
  7. May 13, 2014 #6

    Fredrik

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    I'm pretty sure numbers never decrease. :wink: But ##\epsilon## is an arbitrary positive real number here, so it could certainly be an arbitrary term in a decreasing sequence that converges to 0.

    Since ##\epsilon## is a number you choose, the possibility that M is that specific matrix is not a concern. (If M is that matrix, just change your choice of ##\epsilon##).

    What you need to show is that for all 2×2 matrices M and all ##\varepsilon>0##, there's a 2×2 matrix M' with non-zero determinant such that ##d(M',M)<\varepsilon##. I guess the idea here is that if you choose a small enough ##t>0## and define ##M'=M+tI##, then we will have ##d(M',M)<\varepsilon##. (I haven't tried it, and I'm going to bed now).

    (economicsnerd's ##\epsilon## is my t).
     
  8. May 14, 2014 #7
    Fix your matrix ##M = \begin{pmatrix} a & b\\ c & d \end{pmatrix}##, and notice that ##M+\epsilon I \longrightarrow M## as ##\epsilon \longrightarrow 0##.

    If we can show that the determinant of ##M+\epsilon I## is nonzero for small enough nonzero ##\epsilon##, then we know that matrices with nonzero determinant are dense.* So let's show it.

    The determinant of ##M+\epsilon I## is just ##(a+\epsilon)(d+\epsilon) - bc##, which equals [tex](ad-bc) + (a+d)\epsilon + \epsilon^2[/tex]

    -Exercise: If ##ad-bc\neq 0##, then ##(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0## for small enough nonzero ##\epsilon##. (Though you don't really need to do this one, since ##ad-bc\neq 0## means ##M## already has nonzero determinant.)
    -Exercise: If ##ad-bc=0## but ##a+d\neq 0##, then ##(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0## for small enough nonzero ##\epsilon##.
    -Exercise: If ##ad-bc= a+d = 0##, then ##(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0## for small enough nonzero ##\epsilon##. (This one is basically immediate.)

    *[Indeed, for some ##n\in\mathbb N##, we would know ##M+\frac1k I## has nonzero determinant for every ##k\geq n##. Then ##(M+\frac1k I)_{k=n}^\infty## is a sequence of matrices with nonzero determinant, converging to ##M##.]
     
  9. May 14, 2014 #8
    Great, got it! I'll write it.

    Thank you so much!
     
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