Prove that $A$ is diagonalizable.

Question

Let $A$ be a complex $n\times n$ matrix that satisfies $A^5=I_n$. Prove that A is diagonalizable.

I don't know how to solve this problem. Please help! Thank you.

Answer 1

The minimal polynomial of $A$ divides $X^5-1$. Since $X^5-1$ has distinct roots, then so must the minimal polynomial, so $A$ is diagonalizable.

Answer 2

If $\lambda$ is an eigenvalue of $A$, then $\lambda^5$ is an eigenvalue of $A^5=I_n$. Thus: $\lambda^5=1.$

Let $\lambda_1,...,\lambda_5$ the solutions of the equation $\lambda^5=1.$

These are the eigenvalues of $A$.

Cayley Hamilton gives

$0=(A-\lambda_1 I_n)*...*(A-\lambda_5 I_n)$

thus

$\mathbb C^n=\ker(A-\lambda_1 I_n)\oplus ... \oplus \ker(A-\lambda_5 I_n)$

Let $B_j$ be a basis of $\ker(A-\lambda_j I_n)$, then $B_1 \cup ...\cup B_5$ is a basis of $\mathbb C^n$ consisting of eigenvectors of $A$

Edit: my proof works only in the case $n \ge 5$