# Finding an exponential matrix

How do I find the matrix exponential $e^{tA}$ with

$$A = \left(\begin{matrix} 2 & 8 \\ 0 & 2\end{matrix}\right)$$

The eigenvalue is 2 with multiplicity 2, but it yields only 1 eigenvector {${1, 0}$}, so the matrix isn't diagonalizable. I'm confused what to do. One option is to convert it into a Jordan form, but how do I do that?

Any help is appreciated. Seems like a simple problem, but it's been bugging me for a while.

• $e$ to the power of a (square) matrix is defined similarly to how it is for $e^z = \sum\limits_{n=0}^\infty \frac{z^n}{n!}$ with the caveat that $A^0=I$. Now, note that $A = D+B$, and $A^n = (D+B)^n$ should simplify very nicely for each $n$. (Notice that $B^2 = 0$ and $D$ in this case commutes with $B$ easily) – JMoravitz Apr 22 '16 at 2:25
• The easy way to remember this is to write $tA = 2tI + tN$, where $N$ has zeros on the diagonal and you can see that $N^2$ is the zero matrix. Then, since $2tI$ and $tN$ commute, you have $\exp(tA) = \exp(2tI) \exp(tN)$. The first term you know how to do because it's diagonal. For the second one, use the infinite sum $I + (tN) + (tN)^2/2! + \cdots$... Except since $N^2 = 0$ this infinite sum is just $I + tN$. – Erick Wong Apr 22 '16 at 2:27

## 2 Answers

This one is very simple. You have $$A=2I+X,$$ where $$X=\begin{bmatrix}0&8\\0&0\end{bmatrix}.$$ Because the two matrix commute, we have $e^{t(2I+X)}=e^{2tI}e^{tX}$ (this fails in general for noncommuting matrices). As $X^2=0$, $$e^{tX}=I+tX.$$ Then $$e^{tA}=e^{2tI+tX}=e^{2tI}e^{tX}=\begin{bmatrix}e^{2t}&0\\0&e^{2t}\end{bmatrix}\,\begin{bmatrix}1&8t\\0&1\end{bmatrix} =\begin{bmatrix}e^{2t}&8te^{2t}\\0& e^{2t}\end{bmatrix}.$$

If $D=\begin{bmatrix}2&0\\0&2\end{bmatrix}$ and $N=\begin{bmatrix}0&8\\0&0\end{bmatrix}$, then $A=D+N$, $N^2=0$, and $DN=ND$. Therefore we can apply the binomial theorem to obtain $$A^n=(D+N)^n=D^n+nD^{n-1}N$$ for all $n\geq 1$. Hence $$e^A=\sum_{n=0}^{\infty}\frac{A^n}{n!}=I+\sum_{n=1}^{\infty}\frac{D^n+nD^{n-1}N}{n!}=e^D+N\sum_{n=1}^{\infty}\frac{D^{n-1}}{(n-1)!}=(I+N)e^D=\begin{bmatrix}e^2&8e^2\\0&e^2\end{bmatrix}$$

This sort of trick works whenever you can decompose $A$ into the sum of a diagonal and a nilpotent matrix which commute.

• A small missing detail is how to handle the $t$ in the exponent as well. – JMoravitz Apr 22 '16 at 2:26
• Left as an exercise for the reader... – carmichael561 Apr 22 '16 at 2:27
• Mathematics StackExchange really needs to increase the size of their margins. You can barely fit a proof in there! – Chandler Watson Apr 22 '16 at 2:45