Expected value of the largest singular value of a random matrix with entries in $N (0,1)$

Question

Given a matrix $A \in \mathbb R^{n \times n}$ whose entries are i.i.d. $N(0,1)$, what is the expected value of its largest singular value? Equivalently, what is the expected value of the largest eigenvalue of $A'A$?

Answer 1

If $A$ is a Gaussian random matrix as you describe, then the ensemble of matrices given by $A^TA$ is known as the Wishart ensemble, or the Laguerre ensemble. It has been extensively studied, and you can find information in standard books about random matrix theory.

The average of the largest eigenvalue of $A^TA$ is $4n$. The distribution around the average is given by the Tracy-Widom function. Distribution of large deviations from the mean are also known.

You can start collecting references about largest eigenvalues by looking here:

Large Deviations of the Maximum Eigenvalue in Wishart Random Matrices, by Pierpaolo Vivo, Satya N. Majumdar, Oriol Bohigas, https://arxiv.org/abs/cond-mat/0701371

Answer 2

Largest eigenvalue $x$ of $A'A/n$ converges to scaled/shifted Tracy-Widom distribution. More specifically, the following quantity follows Tracy-Widom distribution.

$$\frac{(x-4) n^{2/3}}{2 \sqrt[3]{2}}$$

The form is given in Theorem 2.15 of "Couillet, Romain, and Zhenyu Liao. 2022. Random Matrix Methods for Machine Learning. Cambridge, England: Cambridge University Press."

You can see that for $n=100$ this distribution gives a good fit to the data:

The mean will be at the following location, where $z\approx -1.2065335745820$ is the mean of standard Tracy-Widom:

$$\frac{2 \sqrt[3]{2} z}{n^{2/3}}+4$$

You can see it approaches 4 at a rate of $n^{-2/3}$, a bit faster than what you see quantities obeying the Central Limit Theorem.

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