For a 2x2 matrix A satisfying $A^k=I$, compute $e^A$ For a 2x2 matrix A satisfying $A^k=I$, compute $e^A$ 
Oh, the exponential of a matrix is: $e^A=\sum_{i=0}^\infty\frac{1}{i!}A^i$

I thought I'd solved the $e^A$ form but I actually did something really silly, now I'm a little stuck.
If $A^k=I$ then $A^{k+1}=A$ and we have a cycle forming, so we will get:
$$\sum^{k-1}_{i=0}\sum^{\infty}_{j=1}\frac{1}{(kj+i)!}A^i$$
(Or something like this form, I don't have paper to hand and just spotted that now)
Is this what the question wants? It doesn't use the 2x2 property.
I believe the answer lies in finding an expression for the inner summation, I can do this for two terms (think of $e+e^{-1}$ all the odd power leave)

If the two eigenvalues are distinct, then $A$ is similar to a diagonal matrix, and its eponential is similar (with the same conjugating matrix) to the exponential of that diagonal matrix, which is found by just exponentiating each entry.
If both eigenvalues are the same, say $\lambda$, then $(A-\lambda I)^2=0$, so $e^A=e^{\lambda I}e^{A-\lambda I}=e^\lambda(I+(A-\lambda I))=e^\lambda(A-(\lambda-1)I)$.
If the matrices are real, then in the latter case, the eigenvalue must be $\pm 1$, in which case $e^A=eA$ or $e^A=e^{-1}(A+2I)$, according to whether it's $+1$ or $-1$. In the former case, the eigenvalues are either imaginary and complex conjugates of each other, or they're $+1$ and $-1$, and you can compute the exponential of the corresponding diagonal matrix accordingly.
I don't think you can say much more without more information; if $k$ is the order of $A$, then you can narrow down the possible roots of unity that the eigenvalues can be, but you still don't know exactly what they are, or what the conjugating matrix is.
