# Dyson’s Brownian Motion

Posted on April 4, 2010 by

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

$M(t) = \frac{A(t) + A^T(t)}{\sqrt{2}}$

Let $\lambda_i(t)$ denote the eigenvalues of $M(t)$. We wish to find the SDE for the $\lambda_i(t)$. In fact,

$d\lambda_i = \sqrt{2}dB_i + \sum_{j\ne i} \frac{1}{\lambda_i-\lambda_j}dt$.

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

By the Markov property, it suffices to do this computation at $t=0$. Suppose we are initially in a basis for which $M$ is diagonal, with diagonal entries $\lambda_i(0)$. By Ito’s lemma, it suffices to find the differential arising from each $B_{ij}$, and add the results together. For $B_{ii}$, we just get $d\lambda_i=\sqrt{2}dB_{ii}$, which is a term we want to have. Now consider off-diagonal entries. It is enough to consider $B_{12}$. Note that the corresponding matrix will have top left 2×2 block

$\left( \begin{array}{cc} \lambda_1(0) & B_{12}(t) \\ B_{12}(t) & \lambda_2(0) \end{array} \right)$

Let $a(t)=\lambda_1(t)+\lambda_2(t)$ and $b(t)=\lambda_1(t)-\lambda_2(t)$. The trace of the displayed matrix is constant, so $da=0$. The determinant of the displayed matrix is $\lambda_1(0)\lambda_2(0)-B^2$, where we have dropped the subscripts on $B$ for convenience. Note that this determinant equals $(a^2-b^2)/4$. Taking differentials, we find that

$d\frac{a^2-b^2}{4}=-2BdB-dt=-dt$

since we are assuming that $B(0)=0$ and working at time $t=0$. Thus we have an ODE for $b^2$, namely $bb'/2=1$ and so

$db=\frac{2}{b}dt$

This is the drift for $db=d(\lambda_1-\lambda_2)$.  Each individual eigenvalue would get just half of this drift (since $\lambda_1+\lambda_2$ is constant in this scenario), so for $d\lambda_1$ we get the drift term

$\frac{1}{\lambda_1-\lambda_2}dt$

which repels the eigenvalues.

We can also answer the question of whether the eigenvalues meet.  Denote $X(t)=\lambda_1(t)-\lambda_2(t)$.  Then, combining Brownian motions, we get

$dX=2dB+\frac{2}{X}dt$

Using Brownian scaling, we get that $2dB(t)=dW(4t)$ for another Brownian motion $W$.  Then setting $s=4t$ and $Y(s)=X(s/4)=X(t)$, we get

$dY=dW+\frac{1}{2Y}dt$

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

I don’t see why when you consider a change of basis such that $M$ is diagonal it turns out that the differential of $\lambda_i$ is $dB_i$. 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 $M$ to be $2x2$. I don’t quite understand why does the $\lambda$ 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 $dB_{ij}$ individually, so the effect of $dB_{ij}$ on the eigenvalues would be determined by the 4 entries whose indices are either $i$ or $j$.

This is a comment, it seems that the argument handles $2x2$ cases but for bigger matrices it would be necessary to conisder more functions than trace ($a$) and $b$ ( 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 $a^2 - b^2$ since $a^2 - b^2 = 4\lambda_1 \lambda_2$ so when you calculate the differential I don’t quite understand why there is the differential of $B_{12}^2$ there, maybe I’m missing something?. ( maybe the functions $a$ and $b$ have a $1/2$ ? )

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

When you calculate the differential of $b^2$ you obtain a term that is $2BdB$ 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 $B_t$ with $B_0=x$ we have $x+B_t$.
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