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)
 A: If $A^k=I$ then $A$ is full rank.  However there is no necessary relation between the rank $n$ and $k$.
E.g. a $2\times 2$ real orthogonal matrix $A$ can have arbitrarily large multiplicative order (think roots of unity), but also $k \gt 0$ can be the order of an arbitrary size matrix (say $n\times n$).
Matrix exponentiation by diagaonlization 
Among the various ways to evaluate matrix exponential, a diagonal matrix $D = \operatorname{diag}(d_1,\ldots,d_n)$ is an easily computed case:
$$ e^D = \operatorname{diag}(e^{d_1},\ldots,e^{d_n}) $$
Also the exponential function conserves similarity relationships because of the power series definition, so if $A = T D T^{-1}$, then:
$$ e^A = T e^D T^{-1} $$
Not every matrix is diagonalizable, but if $A^k = I$ for $k \gt 0$, then $A$ is diagonalizable.  Because $X^k - 1$ has no repeated roots, the minimal polynomial of $A$ must divide $X^k - 1$ and also have no repeated roots.  From this it follows that $A$ is diagonalizable.
Solutions of $A^k=I$ that are real $2\times 2$ matrices
In particular for the $2\times 2$ real matrices, either the minimum exponent $k$ is one, which implies $A=I$ and $e^I = \operatorname{diag}(e,\ldots,e)$, or the minimum exponent $k$ is greater than one (the multiplicative order of $A$).  In the latter case a $2\times 2$ real matrix $A$ will have a conjugate pair of complex eigenvalues that are primitive $k$th roots of unity:
$$e^{\pm 2\pi im/k} = \cos(2\pi m/k) \pm i\sin(2\pi m/k)$$
where integer $0 \lt m \lt k$ is coprime to $k$.
Then for a suitable diagonalizing similarity transformation $T$:
$$ A = T \pmatrix{e^{2\pi im/k}  & 0 \cr
                       0 & e^{-2\pi im/k} \cr } T^{-1} $$
Per the earlier discussion:
$$ e^A = T \pmatrix{e^{e^{2\pi im/k}}  & 0 \cr
                         0 & e^{e^{-2\pi im/k}} \cr } T^{-1} $$
A: 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.
A: At the time I was asked the same question I proposed a bit different solution. If we speak about $\mathbb R^2$ then such an operator is plane isometry so it's reflexion or rotation. In appropriate basis reflexion and rotation have matrices $\begin{pmatrix}
  1 & 0 \\
  0 & -1
 \end{pmatrix}$ and $\begin{pmatrix}
  cos \alpha & -sin\alpha \\
  sin\alpha & cos \alpha
 \end{pmatrix}$, $\alpha = 2\pi/k$. 
As for the reflexion its exponent is easily calculated by substition of its matrix into series definition. For the rotation we may swith to $\mathbb C$ where rotation is just multiplication operator $Az = e^{i\alpha}z$. Its exponent is $e^{e^{i\alpha}}$. Projecting back to $\mathbb R^2$ we get $e^{cos\alpha}\begin{pmatrix}
  cos(sin\alpha) & -sin(sin\alpha) \\
  sin(sin\alpha) & cos(sin\alpha)
 \end{pmatrix}$.
