Notebook 16 – Math 2121, Fall 2020

In this notebook we explore more random matrices and their eigenvalues.

Last time we saw that the eigenvalues of a large orthogonal matrix chosen uniformly at random are, with high probability, evenly spaced complex numbers in the unit circle $\{z\in \mathbb{C}:|z|=1\}$.

Thus, the process of choosing a random eigenvalue of a very large, uniformly random orthogonal matrix is indistiguishable from choosing an element of $\{z\in \mathbb{C}:|z|=1\}$ uniformly at random.

Today we will examine a similar phenomenon for random symmetric matrices.

First let's discuss a few more details about the standard normal distribution.

Here is an intuitive definition of this probability distribution:

• Suppose some students are taking an exam with $q$ different True/False questions.

• On each question, a student has equal chances of answering correctly or incorrectly.

• Students receive $+1$ points for a correct answer, and $-1$ points otherwise.

• The final score on the exam is the sum of these $\pm 1$ points divided by $\sqrt{q}$.

Since we divide by $\sqrt{q}$ instead of $q$, it is possible to receive a score as low as $-\sqrt{q}$ or as high as $+\sqrt{q}$.

Definition: The standard normal distribution is the probability distribution that describes the random scores on this exam when the number of questions is very large.

To observe this empirically, we can simulate the scores for 50000 students on a 10000 question exam.

Now we compare the histogram of scores to a histogram of 50000 samples of the standard normal distribution (computed using a Julia library function).

These plots should look almost the same if our parameters are large enough.

All matrices today have real entries.

A square matrix $S=S^T$ is symmetric if it is equal to its transpose. For example:

All eigenvalues of a symmetric matrix are real numbers. For example:

Compare with the eigenvalues of a random (non-symmetric) square matrix:

Here is a proof of this fact in general.

Assume $S=S^T$ is an $n\times n$ symmetric matrix. This means $S$ has all real entries.

Suppose $0\ne v\in {\mathbb{C}}^n$ is an eigenvector for $S$ with eigenvalue $\lambda \in \mathbb{C}$, so that

$$Sv=\lambda v\text{and}S\overline{v}=\overline{\lambda }\overline{v}.$$

The product ${\overline{v}}^{T}v=\parallel v{\parallel }^2$ is a positive real number since $v\ne 0$.

On the other hand we have both

${\overline{v}}^{T}Sv={\overline{v}}^T(Sv)={\overline{v}}^T(\lambda v)=\lambda {\overline{v}}^{T}v=\lambda \parallel v{\parallel }^2$

and, because $S=S^T$, also

${\overline{v}}^{T}Sv=(S^T\overline{v}{)}^{T}v=(S\overline{v}{)}^{T}v=(\overline{\lambda }\overline{v}{)}^{T}v=\overline{\lambda }{\overline{v}}^{T}v=\overline{\lambda }\parallel v{\parallel }^2.$

These expressions must be equal, and the number $\parallel v{\parallel }^2\ne 0$ is nonzero, so $\lambda =\overline{\lambda }\in \mathbb{R}$.

The set of symmetric $n\times n$ matrices has infinite volume, so it does not make sense to take about a symmetric matrices chosen uniformly at random.

Here is a different way of sampling a random symmetric matrix.

First, we generate a random matrix with entries from the standard normal distribution.

Now to get a random symmetric matrix, we can just form the sum $A+A^T$.

To make later plots nicer, we also divide by the normalizing constant $\sqrt{8n}$.

Here are the (real) eigenvalues of this matrix, in sorted order:

The histogram of these eigenvalues, when $n$ is large, has a surprising shape.

This histogram is the unit semicircle with height rescaled by factor $2/\pi$ so that its area is exactly $1$.

Thus, the eigenvalues of a large random symmetric matrix with entries from the standard normal distribution obey yet another probability law that is neither the uniform distribution nor the normal distribution.

Instead, these random eigenvalues follow what is called the Wigner semicircle distribution.

We can again observe this empirically using Julia's built-in functions for the semicircle distribution: