If H is Hermitian with eigenvalues lambda1 and lambda2 then -H is Hermitian with eigenvaluses -lambda1 and -lambda2. This accounts for identical frequencies for two as for zero positive eigenvalues. --Edwin On Sat, Nov 24, 2012 at 4:16 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Hmm, so I wonder what eigenvalue-sign frequencies would arise if instead of the uniform distribution on [-1,1], one used the standard normal, independently for p, q, r, s.
(Also: is there a simple explanation for why Edwin found almost identical frequencies for two as for zero positive eigenvalues?)
--Dan
On 2012-11-24, at 12:20 PM, W. Edwin Clark wrote:
Consider the case N = 2. A general Hermitian 2 x 2 matrix may be written as H = [r, p + qi] [p - qi, s] where p,q,r,s are arbitrary real numbers.
Let's experiment. Take p,q,r,s to be random real numbers. Since multiplication by a positive scalar c gives eigenvalues of cH of with the same signs we can take the random reals to be in the interval [-1,1].
The eigenvalues of H are
lambda1 = 1/2*s+1/2*r+1/2*((s-r)^2+4*p^2+4*q^2)^(1/2) lambda2 = 1/2*s+1/2*r-1/2*((s-r)^2+4*p^2+4*q^2)^(1/2)
After a million random choices for p,q,r,s in [-1,1] -- using the Maple command rand(-10^10..10^10)()/10.0^10 to generate the random p,q,r,s -- I get the following frequencies:
two positive eigenvalues: 0.0489030 one positive eigenvalue: 0.9021480 zero positive eigenvalues: 0.0489490
So either I'm doing something wrong, or Maple is, or the claim prob(all eigevalues>0) = 2^(-N) is wrong.
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