Calculation matrix exponential 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 
$$
 e^{t \Lambda} = \begin{pmatrix} 
 \exp(t i \omega) & 1 \\
1 & \exp(-ti\omega)
\end{pmatrix}?
$$ 
 A: 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 $\omega.$
A: 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 $(V e^{t\Lambda})V^{-1}$.
A: Here is a result by maple
$$ e^{At}=\left[ \begin {array}{cc} \cos \left( \omega t \right) &\sin \left( \omega t
 \right) \\ -\sin \left( \omega t \right) &\cos \left( \omega t
 \right) \end {array} \right] .$$
A: Several ways you can solve this problem.


*

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

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

*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 .

*Specifically to this question see Differential Equations and Dynamical Systems, Corollary 3, p13.
A: 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 $e^{t\Lambda}=\begin{pmatrix} 
 \exp(t i \omega) & 0 \\
0 & \exp(-ti\omega)
\end{pmatrix}$.
