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$\omega>0$, Then Compute the matrix:

$e^A$, where $A= \begin{bmatrix} 0 & \omega \\ -\omega& 0 \end{bmatrix}$

I have never seen any problem like this. So Please help me solving this. Any link for theoretical reading is welcome.

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    $\begingroup$ Do you know about the relationship between matrix exponentials and diagonalization? $\endgroup$ – Ian Dec 1 '16 at 3:36
  • $\begingroup$ I have learnt about diagonalization but not about exponential.... $\endgroup$ – Arbin Dec 1 '16 at 3:37
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Let $A=XBX^{-1}$, where $B$ is diagonal matrix $[b_1,b_2]$. Then:

$$ e^{A}=\sum_{n=0}^\infty \frac{1}{n!}\left(XBX^{-1}\right)^n=X\left(\sum_{n=0}^\infty \frac{1}{n!}B^n\right)X^{-1}=Xe^BX^{-1}=X[e^{b_1},e^{b_2}]X^{-1} $$

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  • $\begingroup$ How do you know that this happens. Can You provide a link to learn this result, or any references? $\endgroup$ – Arbin Dec 1 '16 at 3:49
  • $\begingroup$ @Arbin It's a pretty standard fact that you should be able to find in any linear algebra reference, in the chapter about eigenvalues. The point is that when you multiply $XBX^{-1}$ with itself $n$ times, you get $X B^n X^{-1}$ because the $X$s cancel out with the $X^{-1}$s. Then if $B$ is diagonal then $B^n$ is simple. Summing them up as in the definition of the matrix exponential gives you an expression for the matrix exponential of $A$. The hard part is actually calculating $X B X^{-1}$ in the first place, which is what I was saying about diagonalization in the comment on the question. $\endgroup$ – Ian Dec 1 '16 at 3:53
  • $\begingroup$ @lan Thank you for this comment. $\endgroup$ – fizis Dec 1 '16 at 4:31
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Hint:

$$e^A=I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots$$

You should calculate $A,A^2,A^3,A^4,A^5,\ldots$ and see if you notice any pattern.

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Here is a long winded way:

Solve the system $\dot{x} = \omega y, \dot{y} = - \omega x$ with initial conditions $(x,y) = (1,0)$ and $(x,y) = (0,1)$.

This gives solutions $t \mapsto (\cos (\omega t), -\sin (\omega t))$ and $t \mapsto (\sin (\omega t), \cos (\omega t))$, and so $e^{At} = \begin{bmatrix} \cos (\omega t) & \sin (\omega t) \\ -\sin (\omega t) & \cos (\omega t) \end{bmatrix}$, and setting $t=1$ gives $e^A = \begin{bmatrix} \cos \omega & \sin \omega \\ -\sin \omega & \cos \omega \end{bmatrix}$.

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