See also: Tao’s blog on Dyson’s Brownian motion

This is an outline of Dyson’s theorem. It isn’t meant to be completely rigorous, and please feel free to expand it. Let be i.i.d. Brownian motions (not starting at 0), and form an matrix with these entries. A second possibility would be to have be complex Brownian motions, but we will stick with the real case. Note that the distribution of is invariant under rotation of coordinates, namely and have the same distribution whenever is an orthogonal matrix. The same is true for , since it is true for each of these matrices individually, and since . This leads to diagonal entries with a coefficient of 2, and off-diagonal entries with a coefficient of . For convenience, we divide the whole matrix by , and call the resulting matrix:

Let denote the eigenvalues of . We wish to find the SDE for the . In fact,

.

In other words, the eigenvalues perform independent Brownian motions with a repelling drift.

By the Markov property, it suffices to do this computation at . Suppose we are initially in a basis for which is diagonal, with diagonal entries . By Ito’s lemma, it suffices to find the differential arising from each , and add the results together. For , we just get , which is a term we want to have. Now consider off-diagonal entries. It is enough to consider . Note that the corresponding matrix will have top left 2×2 block

Let and . The trace of the displayed matrix is constant, so . The determinant of the displayed matrix is , where we have dropped the subscripts on for convenience. Note that this determinant equals . Taking differentials, we find that

since we are assuming that and working at time . Thus we have an ODE for , namely and so

This is the drift for . Each individual eigenvalue would get just half of this drift (since is constant in this scenario), so for we get the drift term

which repels the eigenvalues.

We can also answer the question of whether the eigenvalues meet. Denote . Then, combining Brownian motions, we get

Using Brownian scaling, we get that for another Brownian motion . Then setting and , we get

which is the critical Bessel process, known not to hit 0. Therefore, pairs of eigenvalues don’t meet.

**Questions and Comments**

I don’t see why when you consider a change of basis such that is diagonal it turns out that the differential of is . If you change the basis you get a linear combination of the original brownian motions ( or are you using that sum of two normals is normal and then this is a different brownian motion? ).

- Yes, an orthogonal transformation of an i.i.d. Gaussian vector is again an i.i.d. Gaussian vector.

My second question is when you consider to be . I don’t quite understand why does the appear in the diagonal, nor why these are constants in the calculations latter. I have the impression that the reason is because of Itô’s formula but I don’t see it.

- We are computing the effect of individually, so the effect of on the eigenvalues would be determined by the 4 entries whose indices are either or .

This is a comment, it seems that the argument handles cases but for bigger matrices it would be necessary to conisder more functions than trace () and ( trace with sign ? ). What are these functions? or from this argument we can obtain the result for bigger matrices?.

- See the above comment, the argument is meant to handle matrices of all sizes.

This is another comment, it seems that there is a typo when you calculate the since so when you calculate the differential I don’t quite understand why there is the differential of there, maybe I’m missing something?. ( maybe the functions and have a ? )

- This is a quick post, I wrote “modulo constants”. Please correct the constants if you’re so inclined.

When you calculate the differential of you obtain a term that is but this term is 0 since you are working at time 0, but in the first sentence of the article you mention that the Brownian Motions do not start at 0.

- We switched Brownian motions, so that instead of with we have .

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Posted on April 4, 2010bycarl0mueller0