Nth root of a matrix as an analytic function? Let $A$ be a $k \times k$ invertible matrix over complex numbers.
If it possible to write its nth root as an analytic function (i.e. power series in $A$)?
EDIT: Complex coefficients can be functions of $A$.
Notes
If a matrix $A$ has only one eigenvalue $\lambda$, then it is simple. We take
$$B = \exp\left[\tfrac{1}{n} \log (A ) \right]$$
where we have $B^n = A$. Using Jordan decomposition, we can simplify the logarithm to a polynomial in that matrix (as $(A - \lambda \mathbb{I})$ is nilpotent)
$$\log(A) = \log(\lambda) - \sum_{i=1}^{k} \frac{\left(- \tfrac{A}{\lambda} + \mathbb{I}_k \right)^{i}}{i}.$$
 A: If I am reading this correctly, you are fine with a power series whose (scalar) coefficients depend on the matrix $A$. In this case, it suffices to take a polynomial $p$ that interpolates $\sqrt[n]{x}$, such that for each eigenvalue $\lambda$ with multiplicity $k_\lambda$, the first $k_\lambda-1$ derivatives of $p$ coincide with those of $\sqrt[n]{x}$ (Hermite interpolant). A degree-$k$ polynomial will always do the job.
A: Let $A$ be a $k\times k$ invertible matrix, i.e. in $Gl(k)$. Assume that the segment $[I,A]$ lies in $Gl(k)$.
Let us define
$$
\text{Log}A=\int_{[1,A]} \frac{d\xi}{\xi}=\int_0^1(I-tI+tA)^{-1}(A-I)dt.
$$
It makes sense since $A$ commutes with the denominator inside the integral. The assumption is satisfied in particular whenever $A$ is symmetric invertible  with a nonnegative real part.
Analytic continuation arguments entail
$$
\exp(\text{Log}A)=A\quad \bigl(\exp(\frac{1}{n}\text{Log}A)\bigr)^n=A.
$$
Looking at the Jordan canonical form of $A$, it is not difficult to see that the only thing to be avoided for the above method to work is that eigenvalues should not be negative real numbers. Let $z=a+ib$ be an eigenvalue not in $\mathbb R_-$ in a Jordan block $J_N$ of size $N$, with 1 above the diagonal. Considering the segment $[I_N,J_N]$, we find on the diagonal
$$
(1-t)+tz\notin \mathbb R_-\text{  since $z\notin \mathbb R_-$},
$$
and above the diagonal
$
(1-t)0+t=t.
$
The logarithm formula above works.
A: If all eigenvalues  of $A$ are in the right half plane, there is $\alpha > 0$ such that the circle $|z - \alpha| < \alpha$ contains all the eigenvalues, and the principal branch of $f(z) = z^{1/n}$ is analytic in that circle.  We then have a convergent binomial series
$$ A^{1/n} = \alpha^{1/n} \sum_{k=0}^\infty {{1/n} \choose k} (\alpha^{-1} A - I)^k $$
A: It easy to apply a series to a diagonalizable matrix, and diagonalizable matrices are dense. This should answer your question.
