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let $X^p=M$, where $M, X \in M_n(\Bbb C)$

what's sufficient and necessary conditions for $M$ to have pth roots?

the set of solutions: ${\{A \in M_n(\Bbb C) | A^p=M\}} \ne \emptyset$

also if there any books or references that treat general and arbitrary cases of matrix roots.

Math-Rank-0
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  • $exp:M_n(\mathbb C)\to Gl_n(\mathbb C)$ is surjective and any invertible complex matrix has a logarithm. – Gabriel Romon Mar 12 '17 at 08:39
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    Do you really want the two $n$'s to be the same? – ancient mathematician Mar 12 '17 at 08:42
  • @ancientmathematician: not the same, i changed it to pth roots – Math-Rank-0 Mar 12 '17 at 09:31
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    @marshalcraft i tried to use logarithm like #LeGrandDODOM, but i read $M \in M_n(\Bbb C)$ has a logarithm $iff$ $M$ is inverible, and what about pth roots of singular matrices? – Math-Rank-0 Mar 12 '17 at 09:38
  • I think a good special case to consider might be when $n=4$ and $p=2$ and with the characteristic polynomial of $M$ just $t^4$. In that case I think there are square roots of $M$ if and only if $M=O$ or $M$ has two Jordan blocks of size $2$. For general $n$ and char polynomial $t^n$ we certainly need $p<n$ when $M\not=O$. But ... – ancient mathematician Mar 12 '17 at 13:06

1 Answers1

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i) Let $A\in M_n(\mathbb{C})$. Up to a change of basis, we may assume that $M=diag(U_r,N_{n-r})$ where $U$ is invertible and $N$ is nilpotent.

If $X^p=M$, then $MX=XM$ and $X$ has the form $X=diag(Y_r,Z_{n-r})$ where $Y^p=U,Z^p=N$.

We can choose $Y=\exp(\dfrac{1}{p}\log(U))$ where $\exp(\log(U))=U$. Then the question is: what are the nilpotent matrices that are $p^{th}$ powers ?

ii) The general answer is given by

Psarrakos. On the $m^{th}$ roots of a complex matrix, Electron. J. Linear Algebra 9 (2002) 32–41. cf.

https://repository.uwyo.edu/ela/

$\textbf{Theorem.}$ $A$ has a $p^{th}$ root if and only if the “ascent sequence” of integers $d_1,d_2,\cdots$ defined by $d_i=dim(\ker(A^i))−dim(\ker(A^{i−1}))$ has the property that, for every integer $\nu\geq 0$, no more than one element of the sequence lies strictly between $p\nu$ and $p(\nu+1)$.

iii) The following result is interesting

$\textbf{Theorem}.$ if $A$ is a nilpotent complex matrix, then the set of all $p^{th}$ roots of $A$ is path-connected. cf. De Seguins Pazzis

https://arxiv.org/pdf/1812.02203.pdf