I have the matrix:
$$A=\begin {pmatrix}
0 &-1\\
1 & 0\\
\end{pmatrix}$$
And I want to try to calculate its exponential using this formula $$\ e^{M} = \sum_{k=0}^\infty \frac{1}{k!}\ M^{k}\\$$
I have worked out that $$A^{2} = -I$$ $$A^{3}=-A$$ and $$A^{4}=I$$ where $I$ is the identity matrix. I have then tried to use the fact that the sum will cycle through these matrices to separate the sum and then recombine it into one matrix. However what I get out cannot easily be expressed as a sum. Any help would be greatly appreciated!
Best Answer
Split the sum into odd and even terms
\begin{eqnarray} e^A &=& \sum_{k=0}^{+\infty}\frac{A^{2k}}{(2k)!} + \sum_{k=0}^{+\infty}\frac{A^{2k+1}}{(2k+1)!} \\ &=& \sum_{k=0}^{+\infty}\frac{(-1)^{k}}{(2k)!}I + \sum_{k=0}^{+\infty}\frac{(-1)^{k}}{(2k+1)!}A \\ &=& \cos(1) I + \sin(1)A \\ \end{eqnarray}