Jtdylktuy

Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.

When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= text{initial guess},\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
If I define the number of iterations $k^*$ as
begin{align*}
k^* = min { k : |(I-A) vec x_k - vec b| < epsilon}.
end{align*}
I found that $k^*$ increases as $n$ increases. I know this is intuitively true, but how do I verify this fact rigorously?

If $vec x$ is the actual solution, then we have
begin{align*}
|(I-A) vec x_k - vec b|&=|(I-A) vec x_k - vec b-((I-A)vec x-vec b)|\
&=|(I-A) (vec x_k - vec x)|\
&leq |I-A|| vec x_k - vec x|\
&leq |I-A|frac {|A|^k}{1-|A|}sqrt{n}
end{align*}
Therefore, it's equivalent to define $k^*$ as
begin{align*}
k^* = min { k : |I-A|frac {|A|^k}{1-|A|}sqrt{n} < epsilon}.
end{align*}
Thus, if $epsilon$ is fixed and as $n$ increases, I have to increase $k$ to make sure $|I-A|frac {|A|^k}{1-|A|}sqrt{n} < epsilon$. Therefore, as $n$ increases, $k^*$ increases.