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I got $$ A = \begin{pmatrix} 0 & \omega \\ - \omega & 0 \end{pmatrix}$$ with eigenvalues $\pm i\omega$ and eigenvectors $(-i,1)$ and $(i,1)$. Can I then calculate $e^{tA}$ by $$ e^{tA} = V e^{t \Lambda}V^{-1} $$ where $$ V = \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}; \quad \Lambda = \begin{pmatrix} i \omega & 0 \\ 0 & -i \omega \end{pmatrix} $$ and $$ e^{t \Lambda} = \begin{pmatrix} \exp(t i \omega) & 1 \\ 1 & \exp(-ti\omega) \end{pmatrix}? $$

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The only change you have to do is $$e^{t \Lambda} = \begin{pmatrix} \exp(t i \omega) & 0 \\ 0 & \exp(-ti\omega) \end{pmatrix}.$$ –  Diego Silvera Jan 7 '13 at 23:38
    
wolframalpha.com/input/?i=+{{-i%2Ci}%2C{1%2C1}}.{{exp%28t+i+\omega%29‌​%2C0}%2C{0%2Cexp%28-t+i+\omega%29}}.{{-i%2Ci}%2C{1%2C1}}^{-1}+ –  André Jan 7 '13 at 23:55
    
Wolfram says this is fault. What am I doing wrong ? –  André Jan 7 '13 at 23:56
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In your first comment you need multiply by $V$ and $V^{-1}$, what you wrote is $e^{\Lambda}$. –  Diego Silvera Jan 8 '13 at 0:02
    
And change the ones to zeros. When you exponentiate a diagonal matrix, it remains diagonal. Don't assume that you have to exponentiate all of the entries. –  John Moeller Jan 8 '13 at 1:05
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5 Answers

I strongly encourage you to write out the first, say, eight terms of $$ I + At + \frac{1}{2} A^2 t^2 + \frac{1}{6} A^3 t^3 + \cdots $$ and carefully write out the power series in position 11, then position 12, then position 21, then position 22. Compare these with the power series for the two functions $\cos \omega t$ and $\sin \omega t.$

The trick is that there is a strongly repetitive pattern in $A^n,$ cyclic with period 4 except for the exponent of $\omega.$

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Your calculation of $e^{t\Lambda }$ is incorrect. As it's been pointed out by Diego, $$ e^{t \Lambda} = \begin{pmatrix} \exp(t i \omega) & 0 \\ 0 & \exp(-ti\omega) \end{pmatrix}. $$

Now do $V e^{t\Lambda}$ , and then $(V e^{t\Lambda})V^{-1}$.

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Here is a result by maple

$$ e^{At}=\left[ \begin {array}{cc} \cos \left( \omega t \right) &\sin \left( \omega t \right) \\ -\sin \left( \omega t \right) &\cos \left( \omega t \right) \end {array} \right] .$$

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Several ways you can solve this problem.

  1. By definition; $$ e^{At} = I + At + \frac{1}{2} A^2 t^2 + \frac{1}{6} A^3 t^3 + \cdots $$

  2. By Putzer Algorithm for Finding $e^{At}$, see The Theory of Differential Equations,Classical and Qualitative, p43.

  3. By Diagonalization Process;$$ e^{At} = V e^{\Lambda t}V^{-1}$$ where; $$ e^{\Lambda t} = \begin{pmatrix} \exp(t i \omega) & 0 \\ 0 & \exp(-ti\omega) \end{pmatrix} $$ and $V$ is a Modal matrix .

  4. Specifically to this question see Differential Equations and Dynamical Systems, Corollary 3, p13.

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Your calculation of $e^{t\Lambda}$ is wrong. By definition, for any square matrix $D$, $$ e^D = I + D + \frac{D^2}{2!} + \frac{D^3}{3!} + \ldots $$ In particular, when $D$ is a diagonal matrix (such as your $t\Lambda$), $D,D^2,D^3,\ldots$ are diagonal matrices and hence $e^D$ must be a diagonal matrix too. Furthermore, when $D$ is diagonal, $D,D^2,D^3,\ldots$ are equal to the entrywise powers of $D$; consequently, $e^D$ is also equal to the entrywise exponential of $D$. That is, $\exp\left(\operatorname{diag}(d_1,\ldots,d_n)\right)=\operatorname{diag}(e^{d_1},\ldots,e^{d_n})$. Put $D=t\Lambda$, we see that $e^{t\Lambda}=\begin{pmatrix} \exp(t i \omega) & 0 \\ 0 & \exp(-ti\omega) \end{pmatrix}$.

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