On Nov 22, 2012, at 2:26 AM, Dan Asimov <dasimov@earthlink.net> wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan
That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed. In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2. As Andy pointed out, the probability measure should be invariant under arbitrary unitary transformations, i.e. M -> U M U^(-1). But the Hermitian matrices live in an N^2 dimensional space while the space of unitary matrices has only N(N-1) dimensions. The extra N dimensions correspond to the eigenvalues of M. Wigner had the idea of using the maximum entropy probability distribution, constrained by just two properties: the expectation values of Tr M and Tr M^2. If we want the expectation value of Tr M to be zero, then our probability distribution is simply the Gaussian e^(-Tr M^2) times the unitary-invariant measure. If you marginalize this distribution on just the eigenvalues (i.e. integrate out the unitary transformations) you get, say in the case of N=3 (unnormalized) dP = e^(-E1^2-E2^2-E3^2) (E1-E2)^2 (E2-E3)^2 (E3-E1)^2 dE1 dE2 dE3. It's the product over all eigenvalue pairs -- their differences squared -- that ruins the independence of the eigenvalue distribution. BTW, this very same distribution seems to perfectly model the distribution of Riemann zeta function zeros, but nobody understands why! -Veit
Brent wrote:
Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
2^(-N).
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