2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $e$ 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) $\endgroup$ – JMoravitz Apr 22 '16 at 2:25
  • $\begingroup$ 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 $I + tN$. $\endgroup$ – Erick Wong Apr 22 '16 at 2:27
1
$\begingroup$

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 $$ e^{tA}=e^{2tI+tX}=e^{2tI}e^{tX}=\begin{bmatrix}e^{2t}&0\\0&e^{2t}\end{bmatrix}\,\begin{bmatrix}1&8t\\0&1\end{bmatrix} =\begin{bmatrix}e^{2t}&8te^{2t}\\0& e^{2t}\end{bmatrix}. $$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ A small missing detail is how to handle the $t$ in the exponent as well. $\endgroup$ – JMoravitz Apr 22 '16 at 2:26
  • 1
    $\begingroup$ Left as an exercise for the reader... $\endgroup$ – carmichael561 Apr 22 '16 at 2:27
  • 1
    $\begingroup$ Mathematics StackExchange really needs to increase the size of their margins. You can barely fit a proof in there! $\endgroup$ – Chandler Watson Apr 22 '16 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.