When does the nth root of an affine transformaton matrix exist? I know that we can find the nth root of a lot of transformation matrices, but when can we and when can't we? 
It seems that if we repeatedly apply only one class of transformation, rotation by a bunch of angles for instance, we can find the nth root of that resulting matrix. 
But can we find the nth root of any affine transformation matrix? 
 A: Let $f:x\in\mathbb{C}^n\rightarrow Ux+v\in\mathbb{C}^n$ be a known affine function where $U\in M_n(\mathbb{C}),v\in\mathbb{C}^n$. We seek an affine function $g(x)=Ax+b$ s.t. $g^k=f$ where $g^k$ is the composition $g\circ\cdots\circ g$. The unknowns are $A\in M_n,v\in\mathbb{C}^n$. We obtain for every $x$
$g^k(x)=A^kx+\sum_{i=0}^{k-1}A^i b=f(x)=Ux+v$, that is equivalent to 
$A^k=U,\sum_{i=0}^{k-1}A^i b=v$.
In the sequel, we assume HYP: $U$ has no $\leq 0$ eigenvalue.
Step 1. We choose (for example) $A=\exp(\dfrac{1}{k}\log(U))$, where $\log$ is the principal logarithm. If $spectrum(U)=(\lambda_j)$, then $spectrum(A)=(\mu_j)=(\exp(\log(\lambda_j))/k)$.
Step 2. $\sum_{i=0}^{k-1}A^i$ is invertible iff for every $j$, $\dfrac{\mu_j^k-1}{\mu_j-1}\not= 0$, that is $\dfrac{\lambda_j-1}{\mu_j-1}\not= 0$.  Note that when $\lambda_j=1$, $\mu_j=1$; then $\sum_{i=0}^{k-1}A^i$ is always invertible.
Finally,  $b=(\sum_{i=0}^{k-1}A^i)^{-1}v$.
EDIT. Conclusion. If $U$ has no $\leq 0$ eigenvalues, then $f$ admits at least one $k^{th}$ root. If $U$ has at least one $<0$ eigenvalue, then we must use another logarithm.
