Exponential of matrix

So, I'm wondering if there is an easy way (as in not calculating the eigenvalues, Jordan canonical form, change of basis matrix, etc.) to calculate this exponential $e^{At}$ with $$A=\begin{pmatrix} 0&9 \\ -1&0 \end{pmatrix}.$$

I'd suppose it stumbles around cosines and sines, but I'm not really sure... In fact, something of the kind

$$A=\begin{pmatrix} 3 & 5\cr-5 & -3\end{pmatrix}$$

would also interest me.

• what you are looking for is called the putzers method. i have an answer somewhere explaining this on mss. i don't know how to link it.
– abel
Nov 28, 2014 at 15:44

Let $$A=\pmatrix{0&9\cr-1&0\cr}\ .$$ You can easily check that $$A^2=-9I$$ and so the exponential series gives \eqalign{e^{tA} &=I+tA+\frac{1}{2!}t^2A^2+\frac{1}{3!}t^3A^3+\cdots\cr &=\Bigl(I-\frac{1}{2!}(9t^2I)+\frac{1}{4!}(9^2t^4I)+\cdots\Bigr)+\Bigl(tA-\frac{1}{3!}(9t^3A)+\frac{1}{5!}(9^2t^5A)+\cdots\Bigr)\cr &=(\cos3t)I+\Bigl(\frac{\sin3t}{3}\Bigr)A\ .\cr} For your other matrix you get $B^2=-16I$ and the same method works.

• @Raff and David, I have edited away all the '\\' macro. Nov 28, 2014 at 8:10
• @achillehui That's weird, how could it affect other answers and what was it doing? Nov 28, 2014 at 12:49
• This is a known issue. If you define a macro (using \def or \newcommand) in an answer/question, that definition will not be localized to that particular answer/question. The end result is if you redefine a common macro, the page will randomly pickup one definition and causes unexpected behavior to everyone else. This is what happened to this page and look at this for more discussions on this age-old issue. Nov 28, 2014 at 13:25
• @achillehui Thanks for that. Busy just now but will make sure I check out the links you posted. Nov 28, 2014 at 21:40

If one is allowed to use the values $a$ and $b$ of the eigenvalues, and if $a\ne b$, then one gets

$$\mathrm e^{tA}=\frac{\mathrm e^{at}-\mathrm e^{bt}}{a-b}A+\frac{a\mathrm e^{bt}-b\mathrm e^{at}}{a-b}I,$$

No need here for series expansion or "Jordan canonical form, change of basis matrix, etc.", the substitute being to work directly in the two-dimensional vector space $\mathbb C[A]$, with basis $\{I,A\}$.

To prove the formula above, consider the function $$u(t)=\mathrm e^{At}.$$ The facts that $u'(t)=A\mathrm e^{At}$, $u''(t)=A^2\mathrm e^{At}$, and $A^2-(a+b)A+ab=0$, together yield $$u''(t)-(a+b)u'(t)+abu(t)=0.$$ This (matrix-valued) differential equation is solved by $$u(t)=\mathrm e^{at}U+\mathrm e^{bt}V,$$ for some well chosen (matrices) $U$ and $V$. These are such that $$U+V=u(0)=I,\qquad aU+bV=u'(0)=A.$$ Solving this $2\times2$ Cramer system for $(U,V)$ yields the formula stated at the beginning of this post.

In the end, the method only requires the (seemingly inescapable) step to solve for $(a,b)$ the system

$$a+b=\mathrm{tr}(A),\qquad ab=\mathrm{det}(A).$$

Example 1: If $A=\begin{pmatrix} 0&9 \\ -1&0 \end{pmatrix}$, then $a+b=0$ and $ab=9$ hence one can choose $a=3\mathrm i$ and $b=-3\mathrm i$, then $a-b=6\mathrm i$, $\mathrm e^{at}-\mathrm e^{bt}=2\mathrm i\sin(3t)$ and $a\mathrm e^{bt}-b\mathrm e^{at}=6\mathrm i\cos(3t)$ hence $$\mathrm e^{tA}=\tfrac13\sin(3t)A+\cos(3t)I=\begin{pmatrix} \cos(3t)&3\sin(3t) \\ -\tfrac13\sin(3t)&0\cos(3t)\end{pmatrix}.$$

Example 2: If $A=\begin{pmatrix} 3 & 5\cr-5 & -3\end{pmatrix}$, then $a+b=0$ and $ab=16$ hence one can choose $a=4\mathrm i$ and $b=-4\mathrm i$, then $a-b=8\mathrm i$, $\mathrm e^{at}-\mathrm e^{bt}=2\mathrm i\sin(4t)$ and $a\mathrm e^{bt}-b\mathrm e^{at}=8\mathrm i\cos(4t)$ hence $$\mathrm e^{tA}=\tfrac14\sin(4t)A+\cos(4t)I=\begin{pmatrix} \cos(4t)+\tfrac34\sin(4t)&\tfrac54\sin(4t) \\ -\tfrac54\sin(4t)&\cos(4t)-\tfrac34\sin(4t)\end{pmatrix}.$$

Edit: As noted by @abel in a comment, Putzer algorithm can be used to compute $\mathrm e^{tA}$. For $A$ of size $2\times2$ with eigenvalues $a$ and $b$, the algorithm says that $$\mathrm e^{tA}=r(t)I+s(t)(A-aI),$$ where the (scalar) functions $r$ and $s$ solve the differential system $$r'(t)=ar(t),\quad r(0)=1,\quad s'(t)=bs(t)+r(t),\quad s(0)=0.$$ This yields $r(t)=\mathrm e^{at}$ and, since $a\ne b$, $(a-b)s(t)=\mathrm e^{at}-\mathrm e^{bt}$, thus $$\mathrm e^{tA}=\mathrm e^{at}I+\frac{\mathrm e^{at}-\mathrm e^{bt}}{a-b}\,(A-aI),$$ which is equivalent to the formula we proved above.

• You have a small typo: $u''(t)=A^2 e^{At}$. Nov 28, 2014 at 14:40
• @Artem Indeed. Thanks.
– Did
Nov 28, 2014 at 15:04

These two matrices have a null trace and a positive determinant, so that their Eigenvalues are pure imaginary conjugate. You can infer that the solution will be of the form $C\cos\lambda t+S\sin\lambda t$ (real linear combinations of imaginary exponentials), where $\lambda=\sqrt\Delta$.

Now for $t=0$, $$C\cos\lambda0+S\sin\lambda0=C=e^{A0}=I,$$ and evaluating the first derivative at $t=0$, $$-\lambda C\sin\lambda0+\lambda S\cos\lambda0=\lambda S=Ae^{A0}=A.$$

Thus the final solution

$$I\cos\lambda t+\frac A\lambda\sin\lambda t.$$

We can always find the exponential of a $2 \times 2$ matrix ( with real or complex entries) without calculating eigenvalues or Jordan form.

1) Note that we can always put a $2 \times 2$ matrix in the form $A=hI+A'$ where: $h=\mbox{tr}A/2$, $\mbox{tr}A'=0$, $\mbox{det}A'=\mbox{det}A-\left(\dfrac{\mbox{tr}A}{2}\right)^2$; and $[hI,A']=0$ so that $e^A=e^he^{A'}$.

2) Note that for a traceless matrix $M$ we have: $$M^2=\mbox{det}(M)I$$ so that, given $\theta=\sqrt{\mbox{det}(M)}$, we have: $$\begin{split} e^M&=\sum_{k=0}^\infty\dfrac{M^k}{k!}=\\ &=I+\dfrac{M}{1!}-\dfrac{\theta^2I}{2!}-\dfrac{\theta^3M}{3!\,\theta}+\dfrac{\theta^4I}{4!}+\dfrac{\theta^5M}{5!\,\theta}-\dfrac{\theta^6I}{6!}+\cdots=\\ &=I \left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{M}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\ &=I\cos\theta +\dfrac{M}{\theta}\sin\theta \end{split}$$ 3) So: for every $2 \times 2$ matrix $A=hI+A'$ we have $$e^A=e^h\left(I\cos\theta +\dfrac{A'}{\theta}\sin\theta\right) \qquad \theta=\sqrt{\mbox{det}(A')}$$