Is there a simple proof that the matrix is diagonalizable?

Question

Let

$A^n=D$, where $D$ is a diagonal invertible matrix, and $n\ne0$ is a real number. Is there a simple proof (without involving Jordan canonical form) that the matrix $A$ is diagonalizable?

Answer 1

As $A$ and $D$ commute, eigenspaces of $D$ are stable by $A$. The vector space is therefore a (finite) direct sum of vectors spaces $E_\lambda$ such taht he restriction of $A$ to $E_\lambda$ satisfies $A^n=\lambda. Id$. This restriction is diagonalizable as the polynomial $x^n-\lambda$ has simple roots. Putting together bases of diagonalization of $A$ restricted to the $E_\lambda$ yields a diagonal base for $A$.