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Let $p$ be a positive integer and $M\in Gl_n(\bf C)$

Find some $A\in M_n(\bf C)$ such that $A^p=M$

This problem got me stumped. I can't even deal with $p=2$ (except when $M$ is a positive definite matrix)...

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Gabriel Romon
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1 Answers1

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Here's a method not (explicitly) using Jordan normal form.

We use the following theorem:

For any $A \in GL_n(\Bbb C)$, there is a matrix $X \in M_n(\Bbb C)$ such that $\exp(X) = A$, where $\exp(X) = \sum_{k=0}^\infty \frac 1{k!}X^k$ denotes the matrix exponential.

If we can use this, then we immediately have $$ \left(\exp\left(\frac 1p X\right)\right)^p = \exp(X) = A $$ So that $\exp\left(\frac 1p X\right)$ is a $p$th root of $A$.

That being said, I have not seen a proof of this theorem that does not use Jordan canonical form.

Ben Grossmann
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  • I know this theorem, I even know a proof that doesn't resort to Jordan normal form. How did it occur to you that it would eventually yield a solution ? This is like pulling a rabbit out of your hat. – Gabriel Romon Apr 04 '15 at 05:11
  • Well if I told you, that would ruin the trick :P. The short answer is that when you take a small look into Lie Groups/Algebras as I have, you quickly learn that the exponential map is particularly magical. – Ben Grossmann Apr 04 '15 at 13:20
  • If you could give me a reference for that proof, by the way, I would appreciate it. – Ben Grossmann Apr 04 '15 at 13:23
  • There's also a nice intuition from the complex numbers, over which we define arbitrary powers of a number by $$ z^\alpha = e^{\alpha \ln(z)} $$ – Ben Grossmann Apr 04 '15 at 15:01
  • You can find the proof in there: http://www.amazon.fr/Exercices-math%C3%A9matiques-Oraux-x-ens-alg%C3%A8bre/dp/2842251423 You can find 3 lousy screenshots here http://imgur.com/VRD738X,tuidDln,ygzXEFL#0 . I'll texify it if you can't read French. – Gabriel Romon Apr 04 '15 at 16:04
  • That will do. I have seen something like this, but somehow I forgot how to prove 1 without Jordan Canonical form. Thank you – Ben Grossmann Apr 05 '15 at 00:02