1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the Proportion of Symmetric Matrices that have Positive Determinant?

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    What proportion of 2x2 symmetric matrices with entries belonging to [0, 1] have a positive determinant?


    2. Relevant equations
    [itex]A^{T} = A[/itex]
    If A = [[a, b], [c, d]] Then det(A) = ad - bc. But A is symmetric, so c = b. So det(A) = ad - b^2

    So, in order for A to have a positive determinant, ad > b^2

    3. The attempt at a solution
    I have no idea where to start to get the exact solution. I already did a Monte Carlo simulation which gave the answer .444694. The back of the book gives the solution 4/9, which confirms my monte carlo simulation. How do I get about coming to that fraction for the exact solution?
     
  2. jcsd
  3. Feb 3, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Consider (a, b, d) to be a point in the cube [0,1]x[0,1]x[0,1]. What fraction of the cube satisfies your condition for a positive determinant?
     
  4. Feb 4, 2012 #3
    There are 9 distinct possibilities for relations between a, b, and d.
    a > d > b
    a > d < b
    a > d = b
    a < d > b
    a < d < b
    a < d = b
    a = d > b
    a = d < b
    a = d = b

    If a > d > b then ad > (b^2)
    If a > d = b then ad > (b^2)
    If a < d > b then ad > (b^2)
    If a = d > b then ad > (b^2)

    The condition fails for the other relations between a, b, and d. Therefore the proportion of 2x2 symmetric matrices with a positive determinant (with real entries) is 4/9.
     
  5. Feb 4, 2012 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You said "if a < d > b then ad > (b^2)" but what if a=0, d=1, and b=1/2? In any case, this isn't the right approach anyway.

    I suggest doing an integral to calculate the volume of the region satisfying ad>b2.
     
  6. Feb 5, 2012 #5
    say we have (x, y, z), then xz > y^2. Then xz - y^2 > 0.
    So we want the area under the function f(x, y, z) = xz -y^2, but only when f is positive (right?)

    [itex]\iiint\limits_D xz - y^2 dzdydx = \frac{-1}{12} \hspace{1cm} D = [0,1]\times[0,1]\times[0,1][/itex]

    Okay...that gives me the signed volume, but I only wanted the positive volume. I need to somehow get rid of the negative parts of this volume calculation. I'm not sure how to proceed.
     
  7. Feb 5, 2012 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Just integrate 1 and use the limits to satisfy the condition.
     
  8. Feb 5, 2012 #7
    [itex]\int\limits_0^1\int\limits_0^1\int\limits_0^{y^2/z}dxdydz[/itex]

    doesn't give a meaningful answer, because the last antidifferentiation makes you take ln(0).

    I know 0 < y2 < x*z < 1, how do I represent that as the limits in my triple integral?
     
  9. Feb 5, 2012 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You have xz > y2, so ##y < \sqrt{xz}##, so for a given x and z, y has to be between 0 and ##\sqrt{xz}##.
     
  10. Feb 5, 2012 #9
    Thank you so much for your help. I really appreciate it. I understood how to get to the solution before your final post, but for some reason I posted something that didn't make any sense...perhaps it was too late. You've helped me so that I can solve similar types of problems when I encounter them in the future.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the Proportion of Symmetric Matrices that have Positive Determinant?
Loading...