This is from example 35.1 in Numerical Linear Algebra by Trefethen and Bau.
Suppose that $ B $ is a 200 x 200 real matrix with entries i.i.d from $ \mathcal{N}(0,\sigma^2) $ where $ \sigma = 0.5/\sqrt{200} $. Let $ A = 2I + B $. The eigenvalues of $ A $ are approximately uniformly distributed in the disk of radius 1/2 centered at $ z = 2 $ in the complex plane. The matrix $ I -A/2 $ is then a random matrix scaled so that its spectrum approximately fills the disk of radius 1/4 centered at $ z = 0 $. The authors then claim that this justifies the following: $ \|(I-A/2)^n\|_2 \approx 4^{-n} $. Why is this?
It seems like this should be relatively straightforward, and I must be missing something obvious. In attempting to figure this out, I've come across the formula: $\lim_{n \to \infty} \|C^n\|^{1/n} = \rho(C) $, where $ \rho $ is the spectral radius of $ C $. However, the claim as stated applies to $ n = 1,2,3,\ldots $.
Thanks!
[Edit] In particular, I don't see why the knowledge that the eigenvalues of $ I - A/2 $ are approximately evenly spread over the disk of radius 1/4 centered at $ z = 0 $ implies that $ \|(I-A/2)^n\|_2 \approx 4^{-n} $.
After some investigation, I've concluded that the claim is false.
It should be that $ \|(I-A/2)^n\|_2 \approx 2(4^{-n})$. I don't want to get too far into random matrix theory, but apparently we have that for large $ n $ and $ X_{ij} \sim \mathcal{N}(0,\sigma^2) $, $ \|X\|_2 \approx 2 \sigma \sqrt{n} $ and $ \rho(X) \approx \sigma \sqrt{n} $. Theorem 1.1 in this paper justifies the first approximation. The second approximation is justified because the eigenvalues of $ X $ are distributed uniformly across the disk of radius $ \sigma \sqrt{n} $, so the largest eigenvalue in absolute value will likely be near the edge of the disk. Combining, we have that $ \|X\|_2 \approx 2\rho(X) $.
The random matrix $ I - A/2 $ has the same eigenvalue distribution as $ X $ where $ X_{ij} \sim \mathcal{N}(0, 0.25/\sqrt{200}) $. The $n$th power of $ X $ has eigenvalues spread roughly uniformly over the disk of radius $ \rho(X)^n $, so $ \|(I-A/2)^n\|_2 \approx 2(4^{-n}) $.