If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.

Solution 1:

That passage was written by a user with a Japanese user name "TakuyaMurata". I suppose what he meant was actually this: suppose $F$ is an algebraically closed field and $n>0$ is not an integer multiple of $\operatorname{char}(F)$. If $A$ is invertible and $A^n$ is diagonalisable over $F$, then:

1. $f(x)=(x^n-\lambda_1)\cdots(x^n-\lambda_k)$ annihilates $A$, where $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues of $A^n$.
2. Since the $\lambda_i$s are distinct, $x^n-\lambda_i$ and $x^n-\lambda_j$ have not any common root when $i\ne j$. As $g(x):=x^n-\lambda_i$ and $g'(x)=nx^{n-1}$ also have no common roots (because $\lambda_i\ne0$ and $n\ne0$ --- we have used the assumptions that $A$ is invertible and $n$ is not an integer multiple of $\operatorname{char} F$ here), we see that $f$ has not any multiple roots.
3. As the minimal polynomial $m(x)$ of $A$ divides $f(x)$, $m$ also has not any multiple roots.
4. Therefore, $m$ is a product of distinct linear factors (note that $m$ splits because $F$ by assumption is algebraically closed). Hence $A$ is diagonalisable over $F$.