If J is a Jordan matrix. How to prove that the elements on the principle diagonal of $ e^J$ are exponential function of elements on the principle diagonal of $J$?
If it is two dimensional, then I can prove by calculating it. I have trouble proving it in general n $\times$ n case. Could anyone kindly help? Thanks!
For square matrices, we can sometimes explicitly calculate the infinite sum if there is a clear pattern to $$ e^A = I + A + \frac{1}{2} A^2 + \frac{1}{6} A^3 + \frac{1}{24} A^4 + \cdots $$
A Jordan block is a square (sub)matrix with all the same element $\lambda$ on the main diagonal and then all 1's on the immediate upper off-diagonal. That is, we can write $$ J = \lambda I + N, $$ where $N$ is all zeroes except for the immediate off-diagonal. Since $I$ and $N$ commute, so do $\lambda I$ and $N$ commute.
The main analysis theorem is that, whenever square matrices $A,B$ commute, then $$ e^{A+B} = e^A e^B = e^B e^A. $$ You should familiarize yourself with this theorem; for one thing, find a proof in your book.
Apply to $A = \lambda I $ and $B = N.$ What is $e^N?$ Please do the 2 by 2 and 3 by 3 cases by hand, actually 4 by 4 as well:
$$ N = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) $$
Finally, what is $e^{\lambda I}?$
