## Compute the exponential of a Jordan block

Let $N$ be an $n \times n$ matrix over $\mathbb{C}$ with 1 on the first superdiagonal and 0 elsewhere. Compute $\mathsf{exp}(N\alpha)$. Next, suppose $J$ is an $n \times n$ Jordan block with eigenvalue $\lambda$, and compute $\mathsf{exp}(J\alpha)$.

First, given a parameter $0 \leq t \leq n$, we define a matrix $N_t$ as follows: $N_t = [a_{i,j}]$, where $a_{i,j} = 1$ if $j = i+t$ and $0$ otherwise. (That is, $N_t$ is the $n \times n$ matrix with 1 on the $t$th superdiagonal and 0 elsewhere, and $N = N_1$.)

Lemma: $N_1^k = N_k$ for $k \in \mathbb{N}$. Proof: We proceed by induction. Certainly the result holds for $k = 0$ and $k = 1$. Now suppose it holds for some $k$. We have $N_1(K+1) = N_1N_1^k$ $= N_1N_k$ $= [a_{i,j}][b_{i,j}]$ $= [\sum_\ell a_{i,\ell}b_{\ell,j}]$. Consider entry $(i,j)$: $\sum_\ell a_{i,\ell}b_{\ell,j}$. If $\ell \neq i+1$, then $a_{i,\ell} = 0$. So this sum is $a_{i,i+1}b_{i+1,j}$. Now if $j \neq i+1+t$, then the whole sum is 0; otherwise, it is 1. So in fact $N_1^{k+1} = N_{k+1}$. $\square$

Now $\mathsf{exp}(N\alpha) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!}(\alpha N)^k$ $= \sum_{k \in \mathbb{N}} \dfrac{1}{k!} \alpha^k N_k$. As we saw in this previous exercise, powers of $N$ past the $n-1$th are all zero; so in fact we have $\mathsf{exp}(N\alpha) = \sum_{k=0}^{n-1} \dfrac{1}{k!} \alpha^k N_k$. Note that the indices of nonzero entries of the $N_k$ are mutually exclusive. So $\mathsf{exp}(N\alpha)$ is the $n \times n$ matrix whose $(i,j)$ entry is $\dfrac{\alpha^{j-i}}{(j-i)!}$ if $j \geq i$ and 0 otherwise.

Now let $J$ be the $n \times n$ Jordan block with eigenvalue $\lambda$. Note that $J\alpha = I\alpha + N\alpha$; using this previous exercise, we have $\mathsf{exp}(J\alpha) = e^\alpha \mathsf{exp}(N\alpha)$. (Where $N$ is as above.)