# Matrix exponential for Jordan canonical form

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.

$e^{tX}(x) = \sum_{k=0}^{m} \frac{t^k N^k}{k!} \left(e^{t \lambda_1} \alpha_1 v_1+..+e^{t \lambda_n} \alpha_n v_n \right).$

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.

• Did you mean for $D$ to be diagonalisable or diagonal? If the latter is connecting to a Jordan canonical form of $X$, then in general you need a similarity transformation to get there. – hardmath Sep 25 '15 at 18:50
• diagonalisable. – user167575 Sep 25 '15 at 18:52
• Are we also assuming $D$ and $N$ commute? – hardmath Sep 25 '15 at 19:12
• @hardmath they do, yes. – user167575 Sep 25 '15 at 19:34
• @user167575 If you have read the paper in your link carefully, you'll see that he uses the term "Jordan decomposition" to mean what is usually called "Jordan-Chevalley decomposition" in the literature. Although I don't understand why do you insists in using an overloaded and more ambiguous term, the choice is yours and I'll respect that. – user1551 Sep 27 '15 at 12:12

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 $$0$$.
• @user167575 : If you are using only eigenvectors of $D$, then none of the nilpotent terms are present. For a general vector $x$, you have to expand in terms of the cyclic vectors and their orbits. – DisintegratingByParts Sep 25 '15 at 18:48
• I don't fully understand we have $e^{tX}(x)=e^{tN}e^{tD}.$ Now since $D$ is diagonalisable, there is a full system of eigenvectors with respect to which we can represent $x$ the way I showed you. Then w.r.t. this eigenvectors we have $e^{tD}(v_i) = e^{t \lambda_i} v_i$ and this is how I got my equation (I also expanded $e^{tN}$ but where exactly am I wrong)? – user167575 Sep 25 '15 at 18:53
• @user167575 : Oops. Sorry. The eigenvectors of $D$ include all of the cylic vectors. Yes, that's correct. I had it in my mind that you were asking about the eigenvectors of the original matrix. And, yes, you can see it from what I've written for you because of the exponential property. – DisintegratingByParts Sep 25 '15 at 18:57