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

[Fibo's Rabbit]

Homework Statement

What proportion of 2x2 symmetric matrices with entries belonging to [0, 1] have a positive determinant?

Homework Equations

$A^{T} = A$
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

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?

 

[vela]

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?

 

[Fibo's Rabbit]

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.

 

[vela]

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.

 

[Fibo's Rabbit]

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?)

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

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.

 

[vela]

Just integrate 1 and use the limits to satisfy the condition.

 

[Fibo's Rabbit]

$\int\limits_0^1\int\limits_0^1\int\limits_0^{y^2/z}dxdydz$

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?

 

[vela]

You have xz > y², so $y < \sqrt{xz}$, so for a given x and z, y has to be between 0 and $\sqrt{xz}$.

 

[Fibo's Rabbit]

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.