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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 $A$.


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)

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

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up vote 3 down vote accepted

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.

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Yes, I'm fine with coefficients depending on the matrix. I don't know why I overlooked this solution. – Piotr Migdal Feb 24 '13 at 20:19
However, it is not as simple - I cannot assume that the matrix is diagonalizable; so any function which just maps eigenvalues to their roots won't work. Take as a counterexample $A = [[1, 1], [0, 1]]$ and $f(z)=z$ (sure, another polynomial works for this $A$). – Piotr Migdal Feb 24 '13 at 20:30
The multiplicity of the eigenvalue $1$ is $2$, so you need a polynomial that matches $g(z)$ and $g'(z)$ in $z=1$, where $g(x)=\sqrt[n]{x}$. For some more detail on this approach, you can check Higham's Functions of matrices, SIAM Press 2008. – Federico Poloni Feb 24 '13 at 21:08
Sorry - I meant to add Chapter 1, but I pressed "enter" too quickly. – Federico Poloni Feb 24 '13 at 21:11
@Federico So, you have my thanks in – Piotr Migdal May 8 '13 at 8:54

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.

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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 $$ A^{1/n} = \alpha^{1/n} \sum_{k=0}^\infty {{1/n} \choose k} (\alpha^{-1} A - I)^k $$

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It was my initial thought, but in my case I cannot make such an assumption. – Piotr Migdal Feb 25 '13 at 0:40

It easy to apply a series to a diagonalizable matrix, and diagonalizable matrices are dense. This should answer your question.

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