Exponentiation of a $2\times 2$ matrix

Question

We know:

$$\exp(At)=I+ \sum^{\infty}_{n=1}\frac{A^nt^n}{n!}$$

Here $$A= \begin{pmatrix} 0 & 1 \\ -w^2 & 0\end{pmatrix}$$ is a $2\times 2$ matrix, $I$ is identity matrix.

How to show: $$\exp(At)=I\cos(wt) + A \sin(wt)/w$$

Answer 1

One option is to solve this by diagonalization. Here's another option:

Note that $$ A^2 = -w^2I $$ Thus, we have $$ A^n = \begin{cases} (-1)^{n/2}w^{n}I& n \text{ is even}\\ (-1)^{(n-1)/2}w^{n-1}A& n \text{ is odd} \end{cases} $$ Now, expand the matrix exponential $$ \exp(At)=\sum^{\infty}_{n=0}\frac{A^nt^n}{n!}= \sum^{\infty}_{k=0}\frac{A^{2k}t^{2k}}{(2k)!} + \sum^{\infty}_{k=0}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!} $$ now, reduce each sum, noting the Taylor series for sin and cos.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ We can use Pauli Matrices such that $\ds{{\sf A}t \equiv a + \vec{b}\cdot\vec{\sigma}}$ and $\ds{\expo{{\sf A}t}=\expo{a + \vec{b}\cdot\vec{\sigma}}=\expo{a}\expo{\vec{b}\cdot\vec{\sigma}}}$. Since $\ds{\pars{\vec{b}\cdot\vec{\sigma}}^{2}=\vec{b}\cdot\vec{b}}$, we'll have:

\begin{align} \expo{\vec{b}\cdot\vec{\sigma}}&=\sum_{n\ =\ 0}^{\infty} {\pars{\vec{b}\cdot\vec{\sigma}}^{n} \over n!}\ =\ \overbrace{\sum_{n\ =\ 0}^{\infty} {\pars{\vec{b}\cdot\vec{b}}^{n} \over \pars{2n}!}} ^{\ds{\color{#c00000}{\cosh\pars{\root{\vec{b}\cdot\vec{b}}}}}}\ +\ \overbrace{\bracks{\sum_{n\ =\ 0}^{\infty} {\pars{\vec{b}\cdot\vec{b}}^{n} \over \pars{2n + 1}!}}} ^{\ds{\color{#c00000}{\sinh\pars{\root{\vec{b}\cdot\vec{b}}} \over \root{\vec{b}\cdot\vec{b}}}}} \ \vec{b}\cdot\vec{\sigma} \\[5mm]&=\cosh\pars{\root{\vec{b}\cdot\vec{b}}} +{\sinh\pars{\root{\vec{b}\cdot\vec{b}}} \over \root{\vec{b}\cdot\vec{b}}}\, \vec{b}\cdot\vec{\sigma} \end{align}

In the present case: \begin{align} {\sf A}t&=\pars{\begin{array}{rr}0 & 1 \\ -w^{2} & 0\end{array}}t ={1 - w^{2} \over 2}\pars{\begin{array}{rr}0 & 1 \\ 1 & 0\end{array}}t +{1 + w^{2} \over 2}\,\ic\pars{\begin{array}{rr}0 & -\ic \\ \ic & 0\end{array}}t \\[5mm]&={1 - w^{2} \over 2}\,t\sigma_{x} +{1 + w^{2} \over 2}\,\ic t\sigma_{y} =0 + \pars{{1 - w^{2} \over 2}\,t,{1 + w^{2} \over 2}\,\ic t,0}\cdot \pars{\sigma_{x},\sigma_{y},\sigma_{z}} \\[5mm]&\imp\ \left\{\begin{array}{rcl} a & = & 0 \\[2mm] \vec{b} & = & {1 - w^{2} \over 2}\,t\,\hat{x} + {1 + w^{2} \over 2}\,\ic t\,\hat{y} \end{array}\right. \end{align} and $\ds{\vec{b}\cdot\vec{b}= {\pars{1 - w^{2}}^{2} \over 4}\,t^{2} -{\pars{1 + w^{2}}^{2} \over 4}\, t^{2}=-w^{2}t^{2}}.\quad$ $\boxed{\ds{{\tt\mbox{Note that}}\ \color{#c00000}{{\sf A}t = \vec{b}\cdot\vec{\sigma}}}}$.

Then, \begin{align}&\color{#66f}{\large\expo{{\sf A}t}} =\expo{0}\bracks{\cosh\pars{\verts{wt}\ic} +{\sinh\pars{\verts{wt}\ic} \over \verts{wt}\ic}\,\vec{b}\cdot\vec{\sigma}} =\cos\pars{wt} + {\sin\pars{wt} \over wt}\,{\sf A}t \\[5mm]&=\color{#66f}{\large\cos\pars{wt} + {\sin\pars{wt} \over w}\,{\sf A}} \end{align}

Answer 3

here is another way to verify $X = e^{At} = I \cos(\omega t) + \frac{1}{\omega} \sin( \omega t) \ A$ by showing that $e^{At}$ is the unique solution to the initial value problem $\frac{dX}{dt} = AX$ and $X = I$ at $t = 0.$

we will use the fact $A^2 = -\omega^2 I.$

$\frac{dX}{dt} - AX= -\omega \sin(\omega t) I + \cos( \omega t) A - A[I\cos(\omega t t) + \frac{1}{\omega} \sin(\omega t)A] = -\frac{1}{\omega} \sin(\omega t)[A^2 + \omega^2 I] = 0.$ this establishes the result.