Equality with matrix exponential, sine and cosine

Question

For a square matrix $X$, $e^X$ is given by the power series $$e^X = \sum_{k=0}^\infty{1 \over k!}X^k.$$

Let $$A=\begin{bmatrix}0&1\\-w^2&-a_0\\\end{bmatrix}.$$ Verify that $$e^{At}=\frac{e^{-\lambda t}}{\xi}\{(\xi\cos(\xi t)+\lambda\sin(\xi t))I+A\sin(\xi t)\},$$ where $\lambda=\frac{a_0}{2}$ and $\xi=\left(w^2-\frac{a_0^2}{4}\right)^\frac{1}{2}$.

How should I attack this?

Answer 1

Some (hopefully useful) preliminaries

1. In general, if $A$ is diagonalizable then $A = U\Lambda U^{-1}$ with $U$ the matrix containing the eigenvectors as columns and $\Lambda$ a diagonal matrix formed with the eigenvalues $\Lambda = {\rm diag}\{\lambda_1,\lambda_2,\cdots \}$. Furthermore

   $$ A^k = \underbrace{(U\Lambda U^{-1}) \cdots (U\Lambda U^{-1})}_{k ~{\rm times}} = U\Lambda^k U^{-1}\tag{1} $$

2. Taking this into account

   $$ e^{t A} = \sum_{k=0}^{+\infty}\frac{t^k }{k!}A^k \stackrel{(1)}{=} \sum_{k=0}^{+\infty}\frac{t^k }{k!}U\Lambda^kU^{-1} = U\left(\sum_{k=0}^{+\infty}\frac{t^k }{k!}\Lambda^k\right)U^{-1} = Ue^{t\Lambda}U^{-1} \tag{2} $$

   with $e^{t\Lambda} = {\rm diag}\{e^{t\lambda_1},e^{t\lambda_2},\cdots \}$

Now the problem

In this case, diagonalizing the matrix $A$ is pretty straightforward with, with

$$ \Lambda = \left(\begin{array}{cc} -\lambda - i\xi & 0 \\ 0 & -\lambda + i\xi \end{array}\right) ~~~\mbox{and}~~~ U= \left(\begin{array}{cc} (-\lambda + i\xi)/\omega^2 & (-\lambda - i\xi)/\omega^2 \\ 1 & 1 \end{array}\right) $$

The inverse of $U$ can also be obtained

$$ U^{-1}= \frac{1}{2i\xi}\left(\begin{array}{cc} \omega^2 & \lambda + i\xi \\ -\omega^2 & -\lambda + i\xi \end{array}\right) $$

Evaluating into Eq. (2) we get

\begin{eqnarray} e^{t\Lambda} &=& Ue^{t\Lambda}U^{-1} \\ &=& \frac{e^{-\lambda t}}{\xi}\left( \begin{array}{cc} \lambda \sin (t \xi )+\xi \cos (t \xi ) & \underbrace{\omega^{-2}\left(\lambda ^2+\xi ^2\right)}_{\color{red}{1}} \sin (t \xi ) \\ -\omega ^2 \sin (t \xi ) & \xi \cos (t \xi )\underbrace{-\lambda}_{\color{red}{\lambda - 2\lambda = \lambda -a_0}} \sin (t \xi ) \end{array} \right) \\ &=& \frac{e^{-\lambda t}}{\xi}\left( \begin{array}{cc} \lambda \sin (t \xi )+\xi \cos (t \xi ) & \sin (t \xi ) \\ -\omega ^2 \sin (t \xi ) & \xi \cos (t \xi )+\lambda \sin (t \xi ) -a_0\sin(t \xi) \end{array} \right) \\ &=& \frac{e^{-\lambda t}}{\xi}\left\{(\lambda \sin (t \xi )+\xi \cos (t \xi )) \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) + \sin (t\xi) \left( \begin{array}{cc} 0 & 1 \\ -\omega^2 & -a_0 \end{array} \right)\right\} \\ &=& \frac{e^{-\lambda t}}{\xi}\left\{(\lambda \sin (t \xi )+\xi \cos (t \xi )) I + \sin (t\xi) A\right\} \end{eqnarray}