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 tth 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-1th 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.)

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