Let $X$ be a real $n \times n$ matrix, then there is a Jordan decomposition such that $X = D+N$ where $D$ is diagonalisable and $N$ is nilpotent.
Then, I was wondering whether the following is correct.
$$ e^{tX}(x) = \sum_{k=0}^{m} \frac{t^k N^k}{k!} \left(e^{t \lambda_1} \alpha_1 v_1+\cdots+e^{t \lambda_n} \alpha_n v_n \right).$$
Here $x = \sum_{i=1}^{n } \alpha_i v_i$ and $v_i$ are the eigenvectors of the diagonalisable matrix, $\lambda_i$ are the eigenvalues of $D$ and $m$ is the degree up to which $N^k$ is still non-zero.
Is this correct or am I doing something wrong? Cause I could not find a general equation for this matrix exponential, so I tried my best. (Thus, I am only asking for a verification or correction of this answer.)
If anything is unclear, please let me know.
Best Answer
The linear operators $D$ and $N$ commute, which means that the exponential properties apply: $$ e^{t(D+N)} = e^{tD}e^{tN} = e^{tN}e^{tD}. $$ The exponential of a diagonal is easy enough to compute in any basis for which $D$ is diagonalized. Nilpotent matrices act like higher order differentials when plugging into a power series $F(z)$. For example, if $N$ is nilpotent of order $m$, then $$ F(N)=\frac{F(0)}{0!}I+\frac{F'(0)}{1!}N+\frac{F''(0)}{2!}N^{2}+\cdots+\frac{F^{(m-1)}(0)}{(m-1)!}N^{m-1}. $$ This breaks down nicely when you restrict to one Jordan block. This is because $D=\lambda I$ in that case, which gives $e^{tD}=e^{t\lambda}I$. For example, $$ \exp\left\{t \begin{pmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}\right\} = e^{t\lambda}\begin{pmatrix} 1 & \frac{t}{1!} & \frac{t^{2}}{2!} & \frac{t^{3}}{3!} \\ 0 & 1 & \frac{t}{1!} & \frac{t^{2}}{2!} \\ 0 & 0 & 1 & \frac{t}{1!} \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$ This follows because $$ N = \begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}, N^{2} = \begin{pmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}, N^{3} = \begin{pmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}. $$ By the way, this is all closely related to differential operators. For example, if you want the solutions to $(\frac{d}{dx}-\lambda)^{4}f = 0$, the solutions are combinations of $$ e^{t\lambda},\; e^{t\lambda}\frac{t}{1!},\; e^{t\lambda}\frac{t^{2}}{2!},\; e^{t\lambda}\frac{t^{3}}{3!}. $$ And $(\frac{d}{dx}-\lambda)$ maps the 4th one to the 3rd, the 3rd to the 2nd, etc., and finally maps the first to $0$.