TITLE: $A^k=A$ is diagonalizable over field of order $k$
QUESTION [1 upvotes]: Let $A$ be a square matrix over a field $F$ of order $k$ such that $A^k=A$. Is it true that $A$ is diagonalizable over $F$?
I think it’s true since the minimal polynomial of $A$ over $F$ are products of some factors of $x^k-x$. So the minimal polynomial of $A$ over $F$ is a product of distinct linear polynomials hence $A$ is diagonalizable. However I cannot find any sources about this so I just wanted to verify. Thank you.

REPLY [0 votes]: Question: "I think it’s true since the minimal polynomial of A over F are products of some factors of xk−x. So the minimal polynomial of A over F is a product of distinct linear polynomials hence A is diagonalizable. However I cannot find any sources about this so I just wanted to verify. Thank you."
Question 2: "I'm trying to prove that if a linear operator f is diagonalisable then its minimal polynomial is the product of distinct linear factors."
Remark: If we define an $n \times n$-matrix $A \in M(n\times n,k):=R$ with coefficients in a field $k$ to be diagonalizable iff there is an invertible matrix $P\in R$ with $P^{-1}AP =D$ with $D$ a diagonal matrix, it follows the zero matrix $D(0)$ (the matrix with zeros as coefficients) is a diagonalizable matrix. For any invertible matrix $P$ it follows
$$D(0)=P^{-1}D(0)P.$$
The matrix $D(0)$ has all eigenvalues equal to zero. By definition: The minimal polynomial $f(t)$ of $D(0)$ is the unique monic polynomial generating the ideal $I$ with
$$I:=\{f(t) \in k[t]: f(D(0))=0 \}$$
and $I=(t)$ has $f(t):=t$ as unique monic generator. And this polynomial
has one linear factor.
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