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Assume that $\zeta < 1$ and let $w_d = w_0\sqrt{1-\zeta^2}$.

Show that $$ \exp\left( \begin{array}{cc} -\zeta w_0& w_d \\ -w_d& -\zeta w_0\end{array}\right)t = \left( \begin{array}{cc} e^{-\zeta w_0t}\cos w_dt & e^{-\zeta w_0t}\sin w_dt\\ -e^{-\zeta w_0t}\sin w_dt & e^{-\zeta w_0t}\cos w_dt \end{array} \right) $$

I am not sure how to go about this except to use the matrix exponential formula:

$$e^{At} = I + At + A^2\frac{t^2}{2!} + A^3\frac{t^3}{3!} + ... $$

Can someone please provide me a starting point for this question?

user372834
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2 Answers2

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The hint would be to develop exponential for skew symmetric matrix, and identify terms with trigonometric Taylor series (Rodriguez formula). I explain how to do it:

First, decompose matrix $A$ as a sum of a skew matrix and a multiple of the identity matrix: $$ A = \left[ {\begin{array}{cc} -\xi \omega_0 t & \omega_d t \\ -\omega_d t & -\xi \omega_0 t \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 & \omega_d t \\ -\omega_d t & 0 \\ \end{array} } \right] - \xi \omega_0 t I_2 = \alpha + \beta $$

As both matrices are commutative, i.e. $\alpha \beta = \beta \alpha$, then, one can write: $$ \exp(A) = \exp(\alpha + \beta) = \exp(\alpha) \exp(\beta) $$ (If $AB \neq BA$ then, generally, $\exp(A+B) \neq \exp(A)\exp(B)$)

Developing first exponential: $$\exp(\beta) = I_2 \exp(-\xi \omega_0 t)$$ And the second one: $$\exp(\alpha) = \sum_i \alpha^i/i!$$ As $\alpha$ is a skew matrix: $$\alpha^2 = -(\omega_d t)^2 I_2$$ $$\alpha^3 = -(\omega_d t)^2 \alpha$$ $$\alpha^4 = (\omega_d t)^4 I_2$$ $$\alpha^5 = (\omega_d t)^4 \alpha$$ $$...$$ Which leads to: $$\exp(\alpha) = (1-\frac{(\omega_d t)^2}{2!}+\frac{(\omega_d t)^4}{4!}+\dots)I_2 + (\omega_d t-\frac{(\omega_d t)^3}{3!}+\frac{(\omega_d t)^5}{5!}+\dots)\frac{\alpha}{\omega_d t}$$

Where one can identify $\cos(\omega_d t)$ and $\sin(\omega_d t)$ Taylor series. Finally: $$ \exp(A) = (I_2 \exp(-\xi \omega_0 t))(\cos(\omega_d t)I_2+\sin(\omega_d t)\frac{\alpha}{\omega_d t})$$ $$ \exp(A) = \left[ {\begin{array}{cc} \exp(-\xi \omega_0 t)\cos(\omega_d t) & \exp(-\xi \omega_0 t)\sin(\omega_d t) \\ -\exp(-\xi \omega_0 t)\sin(\omega_d t) & \exp(-\xi \omega_0 t)\cos(\omega_d t) \\ \end{array} } \right] $$ Obtaining the final result

Carlos Campos
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There is a rather easy way to understand it through the (field) isomorphism between the set of matrices :

$$Z=\begin{pmatrix} a & -b \\ b& \ \ a\end{pmatrix} \ \leftrightarrow \ z=a+ib$$

and the set of complex numbers $\mathbb{C}$ with the particular case :

$$\tag{1}\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta)& \ \ \cos(\theta)\end{pmatrix} \ \leftrightarrow \ \cos(\theta)+i \sin(\theta)=e^{i \theta}.$$

(Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$).

Here is how this isomorphism will be helpful. Let us set :

$$\tag{2}\begin{cases}\cos(\theta)&=&\zeta\\\sin(\theta)&=&\sqrt{1-\zeta^2}\end{cases}, \ \text{and thus} \ w_d=w_0 \sin(\theta).$$

for a certain $\theta \in (0,\pi/2)$.

Let $A$ and $B$ be the LHS and RHS matrices, resp.

Using relationships (2), and then isomorphism (1), we have:

$$\tag{3}A=\exp \left[ -tw_0\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta)& \ \ \cos(\theta)\end{pmatrix} \right] \ \ \ \leftrightarrow \ \ \ e^{-tw_0 e^{i \theta}}=e^{-tw_0 (\cos(\theta)+i\sin(\theta))}$$

$$\tag{4}B=e^{-\zeta w_0 t}\left( \begin{array}{cc} \cos(-w_dt) & -\sin(-w_dt)\\ \sin(-w_dt) & \ \ \cos(-w_dt) \end{array} \right) \ \ \ \leftrightarrow \ \ \ e^{-\zeta w_0 t}e^{-i tw_d} = e^{-t w_0 \cos(\theta)}e^{-it w_0 \sin(\theta)}$$

(3) and (4) are identical, thus $A$ and $B$ are identical too.

Remark : We have used in (4) the fact that isomorphism (1) can be extended to functions, in particular to exponentials : if $Z \ \leftrightarrow \ z$, then $\exp(Z) \ \leftrightarrow \ e^z$ (as an immediate consequence of the common series definition).

Jean Marie
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