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
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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)


$$\log(A) = \log(\lambda) - \sum\_{i=1}^{k} \frac{\left(- \tfrac{A}{\lambda} + \mathbb{I}\_k \right)^{i}}{i}.$$



Nth root of a matrix as an analytic function?
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.


