Suppose that I have a collection of matrices $\{ A_i \mid A_i \in \mathcal{M}(n;\mathbb{R}) \text{ for } 1 \le i \le k \}$. Suppose further that I know that each $A_i$ has only complex (non-real) roots of its characteristic polynomial (I'm hesitant to use the word eigenvalue, as there is no corresponding (real) eigenvector), furthermore suppose that all roots have modulus less than 1. Now, several things seem like they ought to be obvious, but I don't know how to prove them. First, it seems that if I consider the image of the standard $(n-1)$-sphere under $A_i$ (i.e., $A_i \sim S^{n-1} \subset \mathbb{R}^n$) that the resulting ellipsoid (i.e., the image) lies "mostly" inside the standard $(n-1)$-sphere. How precise can "mostly" be made and how can one prove the precise statement. If the preceding is true, then it would seem that (so long as the matrices are "independent") products of such matrices must be contractions "almost everywhere" (at least if I consider countably infinite products, in which case I would expect the product to be zero almost everywhere). I can't imagine a universe where these statements are not true, but am struggling to make them precise and construct proofs. Any suggestions or pointers would be most welcome.

In response the thoughtful comments below, let me expand a bit:

In my original posting, I was striving for brevity and was, perhaps too terse – I apologize. The situation in which I am interested is random matrices (appropriately scaled by $1/\sqrt{n}$). I am principally interested in the case of very large $n$ (my ultimate interest is asymptotic in $n$), and I want to understand the geometric consequences of the Circular Law (i.e., I am familiar with the result that the scaled eigenvalues are distributed uniformly in the unit disk). I do understand that the convergence to the Circular Law is only in distribution, so that one may have “many” (in the case of finite $n$) “ill-behaved” eigenvalues. Still, it seems to me that knowing that for large $n$, the majority of the eigenvalues will have complex modulus less than one ought to tell me something about the Lipschitz constant (operator norm) of the random matrix over some “reasonably large portion” of the $(n-1)$-sphere.

Ultimately, I am interested in products of random matrices (the $k$ in my original posting). As the eigenvectors of i.i.d. random matrices are distributed uniformly on the (complex) $(n-1)$-sphere, it seems that I ought to be able to say that, for $k$ sufficiently large, the Lipschitz constant of the product is bounded by $\epsilon$ away from a set of measure $\delta$ - a statement of this form is my ultimate goal.

I do know that the eigenvalues of a $k$-fold product of i.i.d. random matrices converges to the $k^{\mbox{th}}$ power of the uniform distribution. Thus, in terms of complex modulus, the eigenvalues quickly concentrate near zero. But, I would prefer to some sort of estimate of the behavior in terms of each matrix in the product. It seems, maybe, that explicit estimates of this form might be more difficult to produce than I originally thought.

  • $\begingroup$ I guess "has only complex roots" means "no complex root is real"... yes eigenvalue is standard in this context, and quite clear if you say complex eigenvalue (and any real number is a complex number). $\endgroup$
    – YCor
    Aug 21, 2019 at 17:52
  • 1
    $\begingroup$ If you are not assuming normality of your matrices, then the eigenvalues may fail to control the image of the unit ball, sometimes drastically. Consider $A=\left(\begin{matrix} 1 & 10^6 \\ 0 & 1 \end{matrix} \right)$ $\endgroup$
    – Yemon Choi
    Aug 21, 2019 at 23:19
  • $\begingroup$ True that the eigenvalue in your example is one and that in the direction of the major axis of the ellipse that results from applying $A$ to $S^1$ the image greatly exceeds unit modulus, but along the minor axis of the resulting ellipse is almost degenerate. So, it is a very small proportion of the circumference of the $S^1$ where the image exceeds. $\endgroup$ Aug 22, 2019 at 14:06

1 Answer 1


I do not understand the role of $k$ in your question, so I will just ignore it.

If you take your matrix to consist of independent Gaussian entries (normalized by $\sqrt{n}$), this will give a precise meaning to the statements you are trying to make (since you can discuss "almost all" or "a lot" in terms of the probability of the event you care about). These are (real) Ginibre matrices, and a lot is known. For example, it is true that with high probability, all eigenvalues are smaller than $1+o_N(1)$ (google "the circular law"), but there are plenty of singular values larger than $1$ (the top one is at $\sqrt{2}$ if I recall right, google "Wishart matrix"). So in particular, it is not true that the image is contained in the unit ball, even with a relaxed understanding of the notion of "mostly".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.