Asymptotic Properties of Jordan Block 2-norm Let $A = SJS^{-1}$ denote the Jordan decomposition of the matrix $A\in\mathbb{R}^{n\times n}$, where $J = \textrm{blkdiag}(J_1,\ldots,J_m)$, where each Jordan block is given by
\begin{equation*}
J_i = \lambda_i I_{n_i} + N_{n_i} \in \mathbb{R}^{n_i\times n_i},
\end{equation*}
with $N_{n_i}$ denoting the nilpotent matrix (i.e., $N_{n_i}^{n_i} = 0$) with ones on the superdiagonal and zeros elsewhere.
I am interested in determining the asymptotic behavior of the two-norm of the Jordan matrix, that is, $\lim_{k\rightarrow\infty} \| J^k \|_{2}$.
To this end, note the binomial expansion for $J_i^k$ as
\begin{align*}
J_i^k = (\lambda_i I_{n_i} + N_{n_i})^k &= \sum_{\ell = 0}^{k} 
\binom{k}{\ell}
(\lambda_i I_{n_i})^{k-\ell} N_{n_i}^{\ell} \\
&= \sum_{\ell=0}^{k} 
\binom{k}{\ell}
\lambda_i^{k - \ell} N_{n_i}^{\ell} \\
&= \sum_{\ell = 0}^{\min(k,n_i)} 
\binom{k}{\ell}
\lambda_i^{k-\ell}N_{n_i}^{\ell} \\
&= 
\binom{k}{0}\lambda_i^k I_{n_i} + 
\binom{k}{1}\lambda_i^{k-1} N_{n_i} + \ldots + 
\binom{k}{n_{i-1}}\lambda_i^{k-n_i + 1}N_{n_i}^{n_i-1} \\
&=
\begin{bmatrix}
 \lambda_i^k & \binom{k}{1}\lambda_i^{k-1} & \binom{k}{2}\lambda_i^{k-2} & \cdots & \cdots & \binom{k}{n_i-1}\lambda_i^{k-n_i+1} \\
 & \lambda_i^k & \binom{k}{1}\lambda_i^{k-1} & \cdots & \cdots & \binom{k}{n_i-2}\lambda_i^{k-n_i+2} \\
 &  & \ddots & \ddots & \vdots & \vdots\\
 &  & & \ddots & \ddots & \vdots\\
 &  & &  & \lambda_i^k & \binom{k}{1}\lambda_i^{k-1}\\
 &  &  &  &  & \lambda_i^k
\end{bmatrix}
\end{align*}
So, this is the structure of each Jordan block $J_i^k$.
Now, the question is how to evaluate or do asymptotic analysis on $\|J^k\|_{2} := \sigma_{\textrm{max}}(J^k) = \sqrt{\rho((J^k)^\intercal J^k)}$.
In the textbook Iterative Methods by Greenbaum, they write the expression
\begin{equation*}
 \|J^k\| \sim \binom{k}{n_i-1}[\rho(J)]^{k-n_i+1}, \quad k\rightarrow\infty,
\end{equation*}
which (vaguely) corresponds to the upper-right element in the above expression for $J_i^k$.
I am just not sure how to make the connection from $J_i^k$ to $\lim_{k\rightarrow\infty} \|J^k\|$.
 A: First of all, note that $J^k = \operatorname{blkdiag}(J_1^k ,\dots, J_m^k)$, so that $\|J^k\| = \max_i\|J_i^k\|$. Also, for sufficiently large $k$, $\|J_i\|$ will necessarily be greatest for an $i$ such that $|\lambda_i| = \rho(J)$ (as a consequence of the fact that $\|J_i^k\|^{1/k} \to |\lambda_i|$). Thus, it suffices to consider the case where $J = J_i$ for such an $i$. With that in mind, I will drop the $i$ subscripts from this point on; note that we will have $\rho(J) = |\lambda|$.
For convenience, write $f(k) = \binom{k}{n-1}[\rho(J)]^{k-n+1}$. In order to justify the asymptotic equivalence, we need the following two facts:

*

*$J^k$ has an entry with absolute value equal to $f(k)$ (namely the upper-right element),

*All other entries of $J^k$ are $o\left(f(k)\right)$ (see the end of my answer for a proof, and for explanation of what $o(\cdot)$ means if that is unfamiliar notation).

With that, let $e_n$ denote the final standard basis vector, i.e. the last column of the identity matrix. We have the following lower-bound on $\|J^k\|$:
$$
\|J^k\|^2 \geq \|J^k e_n\|^2 = \sum_{j=0}^{n-1} \left|\binom{k}{j}\lambda^{k-j} \right|^2 \geq \left|\binom{k}{n-1}\lambda^{k-n+1} \right|^2 
= [f(k)]^2.
$$
For the upper bound, select any $0 < \epsilon < 1$. Let $\|\cdot\|_F$ denote the Frobenius norm. Let $k_0$ be such that $|[J^k]_{p,q}| \leq \frac{\epsilon}{n} f(k)$ for all $1 \leq p,q \leq n$ whenever $k > k_0$. For such $k$, we have
\begin{align}
\|J^k\|^2 & \leq \|J^k\|_F^2 = |[J^k]_{1,n}|^2 + \sum_{p,q; (p,q) \neq (1,n)} |[J^k]_{p,q}|^2 
\\ & = f(k)^2 + \sum_{p,q; (p,q) \neq (1,n)} |[J^k]_{p,q}|^2 
\\ & \leq 
f(k)^2 + \frac{n^2-1}{n^2} \epsilon^2 f(k)^2
\\ & \leq f(k)^2 + \epsilon^2f(k)^2 \leq f(k)^2 + \epsilon f(k)^2 = (1 + \epsilon)f(k)^2.
\end{align}
Thus, for any $0 < \epsilon < 1$ and sufficiently large $k$,
$$
f(k)^2 \leq \|J^k\|^2 \leq (1 + \epsilon)f(k)^2.
$$
Thus, for all such $\epsilon$, we have
$$
1 \leq \lim_{k \to \infty} \frac{\|J^k\|^2}{f(k)^2} \leq 1 + \epsilon,
$$
and since $0 < \epsilon < 1$ was arbitrary this implies that $\lim_{k \to \infty} \frac{\|J^k\|^2}{f(k)^2} = 1$. That is, $\|J^k\|^2 \sim [f(k)]^2$, which means that $\|J^k\| \sim f(k)$.

Proof of fact 2: Every entry besides the top-right entry can be written as $\binom kj \lambda^{k-j}$ for some $j < n-1$. We note that (for $k \geq n-1$)
\begin{align}
\frac{\left|\binom kj \lambda^{k-j}\right|}{f(k)} &= 
\frac{\binom{k}{j}|\lambda|^{k-j}}{\binom{k}{n-1}|\lambda^{k-n+1}|}
= 
|\lambda|^{n-1-j} \cdot \frac{\binom{k}{j}}{\binom{k}{n-1}}
\\ & = 
|\lambda|^{n-1-j} \cdot \frac{k!}{j!(n-j)!} \cdot \frac{(n-1)!(k-n+1)!}{k!}
\\ & = |\lambda|^{n-1-j} \cdot \frac{(n-1)!(k-n+1)!}{j!(k-j)!}
\\ & = 
|\lambda|^{n-1-j}
\frac{(j+1)(j+2)\cdots (n-1)}{(k-n+2) \cdots (k-j+1)(k-j)}
\\ & = 
\frac{|\lambda|^{n-1-j}[(j+1)(j+2)\cdots (n-1)]}{(k-n+2) \cdots (k-j+1)(k-j)}.
\end{align}
We note that the final expression has a constant numerator (relative to $k$) and denominator that tends to $\infty$ as $k \to \infty$. Conclude that $\left|\binom kj \lambda^{k-j}\right|\Big/f(k) \to 0$ as $k \to \infty$, which is to say that $\left|\binom kj \lambda^{k-j}\right| = o(f(k))$.
Alternatively, fact 2 can be proved a bit more quickly if we use the fact that $\binom kj = \Theta(k^j)$, which is to say that $\binom kj \sim ck^j$ for some $c > 0$.
