0

Let $A \in M_n(\mathbb{C})$ be an invertible matrix.

Prove that $A$ is diagonalizable iff $A^{m}$ is diagonalizable for $m \in \mathbb{N}$

I have proved that if $A$ is diagonalizable then $A^{m}$ is diagonalizable.

But I didn't know how to prove that if $A^{m}$ is diagonalizable for $m \in \mathbb{N}$ then $A$ is diagonalizable.

FAF
  • 405
  • Yes. Thank you!! – FAF May 07 '20 at 15:15
  • 1
    Which is the same as your other direction: if $A^m$ is diagonalizable for $m\in N$, then $A$ is diagonalizable. S – Matcha Latte May 07 '20 at 15:15
  • 1
    User767941, I spent two seconds on finding the answer, please, search the site. – Matcha Latte May 07 '20 at 15:16
  • Let's use $\mathbb{R^+}$ instead of $\mathbb{N}$ for the exponent $m$ (there's no real problem in $m$ being a real instead of just a natural). Use the variable change $A^m=B$. Then $A=B^{1/m}$. So, for what you proved, $B$ diagonalizable implies $B^{1/m}$ also diagonalizable, and we have finished, because if it's true for $m\in\mathbb{R}$, then it is for $m\in\mathbb{N}$ ($\mathbb{N}\subset\mathbb{R}$). – Alejandro Bergasa Alonso May 07 '20 at 19:27

0 Answers0