# Calculating the matrix exponential of rotation matrix

I have the matrix:

$$A=\begin {pmatrix} 0 &-1\\ 1 & 0\\ \end{pmatrix}$$ And I want to try to calculate its exponential using this formula $$\ e^{M} = \sum_{k=0}^\infty \frac{1}{k!}\ M^{k}\\$$

I have worked out that $$A^{2} = -I$$ $$A^{3}=-A$$ and $$A^{4}=I$$ where $$I$$ is the identity matrix. I have then tried to use the fact that the sum will cycle through these matrices to separate the sum and then recombine it into one matrix. However what I get out cannot easily be expressed as a sum. Any help would be greatly appreciated!

• You should get a numerical series times $I$ plus a numerical series times $A$. Both these series of numbers should look familiar... Mar 1 '18 at 18:47
• Collect the terms in $A$ and $I$ and factor those matrices out. Do you recognize the resulting series?
– amd
Mar 1 '18 at 18:47

Split the sum into odd and even terms

\begin{eqnarray} e^A &=& \sum_{k=0}^{+\infty}\frac{A^{2k}}{(2k)!} + \sum_{k=0}^{+\infty}\frac{A^{2k+1}}{(2k+1)!} \\ &=& \sum_{k=0}\frac{(-1)^{k}}{(2k)!}I + \sum_{k=0}^{+\infty}\frac{(-1)^{k}}{(2k+1)!}A \\ &=& \cos(1) I + \sin(1)A \\ \end{eqnarray}

• +1 elegant answer..Cant you consider I as $A^0$ ? Mar 1 '18 at 19:04
• @Isham Thanks for the suggestion, you can indeed. I already fixed it Mar 1 '18 at 19:06

You have\begin{align}e^A&=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&-1\\1&0\end{pmatrix}+\frac12\begin{pmatrix}-1&0\\0&-1\end{pmatrix}+\frac1{3!}\begin{pmatrix}0&1\\-1&0\end{pmatrix}+\frac1{4!}\begin{pmatrix}1&0\\0&1\end{pmatrix}+\cdots\\&=\begin{pmatrix}1-\frac1{2!}+\frac1{4!}-\cdots&-1+\frac1{3!}-\frac1{5!}+\cdots\\1-\frac1{3!}+\frac1{5!}-\cdots&1-\frac1{2!}+\frac1{4!}-\cdots\end{pmatrix}\\&=\begin{pmatrix}\cos(1)&-\sin(1)\\\sin(1)&\cos(1)\end{pmatrix}.\end{align}

A sketch:

First consider this matrix as a matrix in $\mathcal M_2(\mathbf C)$, and diagonalise it. Its characteristic polynomial is $\lambda^2+1$, so the eigenvalues are $i,-i$. One finds corresponding eigenvectors: $e_1=(1,1)$ and $e_2=(1,i)$. In this basis, the diagonal matrix $D$ of eigenvalues has exponential: $$\exp D=\exp\begin{pmatrix}\mathrm e^i &0\\0&\mathrm e^{-i}\end{pmatrix}.$$ Now, the change of basis matrix from the canonical basis to basis $(e_1,e_2)$ and its inverse are $$P=\begin{pmatrix}i&1\\1&i\end{pmatrix},\quad P^{-1}=\tfrac12\begin{pmatrix}-i&1\\1&-i\end{pmatrix}.$$ Also, we know that $D=P^{-1}AP,\quad \text{so }\enspace A=PDP^{-1}$and $$\exp A=P\exp(D)P^{-1}=\text{ (some computation) }= \begin{pmatrix}\cos 1&-\sin 1\\\sin 1&\cos 1\end{pmatrix}$$ i.e. the matrix of rotation by 1 rad.

It is straightforward using the formula above to check that $t \mapsto e^{At}$ satisfies the ODE $\dot{X} = AX$, $X(0) = I$. We want to compute $X(1) = e^A$.

Note that $\ddot{X} = -X$, so each entry has the general solution $t \mapsto a \cos t + b \sin t$, so the general solution is $X(t)=(\cos t)C + (\sin t) S$ for some matrices $C,S$.

The initial condition gives $C=I$, and the ODE gives $\dot{X}(t) = -(\sin t) I + (\cos t) S = (\cos t)A + (\sin t) AS$ from which we see that $S=A$.

Hence $X(t) = e^{At} = (\cos t)I + (\sin t) A$. Choosing $t=1$ gives the desired answer.

$${\rm A}^2 = - {\rm I}_2$$

Hence, matrix $${\rm B} := \frac{{\rm A}}{i}$$ is involutory, i.e., $${\rm B}^2 = {\rm I}_2$$. Using Euler's formula,

\begin{aligned} \exp({\rm A}) = \exp \left( i {\rm B} \right) &= \cos \left( {\rm B} \right) + i \sin \left( {\rm B} \right)\\ &= \cos \left( 1 \right) \, {\rm I}_2 + i \sin \left( 1 \right) \, {\rm B}\\ &= \color{blue}{\cos \left( 1 \right) \, {\rm I}_2 + \sin \left( 1 \right) \, {\rm A}}\end{aligned}

where $$\cos \left( {\rm B} \right) = \cos \left( 1 \right) \, {\rm I}_2$$ and $$\sin \left( {\rm B} \right)= \sin \left(1 \right) \, {\rm B}$$ because $${\rm B}$$ is involutory.