# Computing the matrix exponential for a Jordan matrix

How can I compute $$e^{At}$$ where $$A = J_{3}(5)$$? That is,

$$A = \begin{pmatrix} 5 & 1 & 0 \\ 0 & 5 & 1 \\ 0 & 0 & 5 \end{pmatrix}$$

Using this, how can I write down a basis for the space of solutions to the differential system $$dx/dt = Ax$$?

I computed the characteristic polynomial $$(-\lambda + 5)^3$$, which has roots at $$\lambda = 5$$ with multiplicity $$3$$. Hence, we can write

$$e^{At} = \sum_{k=0}^{n - 1}c_{k}A^{k},$$

where the $$c_{k}'s$$ are determined by

$$e^{\lambda_{i}t} = \sum_{k=0}^{n-1}c_{k} \lambda_{k}.$$

So,

$$e^{5t} = c_{0}5 + c_{1}5 + c_{2}5$$

I got stuck from here. I wrote the first sum explicitly as

$$e^{At} = c_0A^{0} + c_1A^{1} + c_{2}A^{2},$$

but I got nowhere. I would really appreciate some help.

Since$$A=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$and since these two matrices commute,$$e^A=\exp\left(\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}\right)\exp\left(\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right).$$Can you take it from here?

• So I can write $e^{5I_{3\times 3}} = I_{3\times 3}e^{5}$ by some property in my book. How do I get rid of the second term? – user663014 Apr 15 at 7:49
• Yes, that is correct. – José Carlos Santos Apr 15 at 7:50
• Sorry, I mean how do I simplify the $\text{exp}\left(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}\right)$ term? Or, would my answer just be $I_{3\times 3} e^{5} \text{exp}\left(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}\right)$? – user663014 Apr 15 at 7:50
• How does this relate to the space of solutions for $dx/dt = Ax?$ – user663014 Apr 15 at 8:03
• $$\exp\left(\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right)=\begin{bmatrix}1&1&\frac12\\0&1&1\\0&0&1\end{bmatrix}.$$ – José Carlos Santos Apr 15 at 8:13

Write $$A$$ as

$$A=S+N= \begin{pmatrix} 5&0&0\\ 0&5&0\\ 0&0&5 \end{pmatrix}+ \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}$$

namely write $$A=S+N$$, where $$S$$ is the semisimple part of $$A$$ and $$N$$ is the nilpotent part of $$A$$. It easy to check that $$[S,N]=0.$$

Now: $$e^A=\sum_\Bbb N\frac{1}{n!}A^n$$

hence

$$e^{A}=e^{S+N}=\sum_{\Bbb N}\frac{1}{n!}(S+N)^n=\sum_{\Bbb N}\sum_{k=1}^n\frac{1}{n!}{{n}\choose{k}}S^kN^{n-k}=\sum_{\Bbb N}\sum_{k=1}^n\frac{1}{k!(n-k)!}S^kN^{n-k}=e^Se^N$$

Since $$N^2=0$$, you have

$$e^{S}=\begin{pmatrix} e^5&0&0\\ 0&e^5&0\\ 0&0&e^5 \end{pmatrix}$$

and

$$e^N=\begin{pmatrix} 1&1&\frac12\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$

then $$e^{A}=e^{S+N}=e^Se^N= \begin{pmatrix} e^5&e^5&\frac{e^5}{2}\\ 0&e^5&e^5\\ 0&0&e^5 \end{pmatrix}$$

• Hey, thanks. How does this relate to the solutions to $dy/dx = Ax?$ – user663014 Apr 15 at 8:08