Let $(A,\Theta)$ be a principally polarized abelian variety over an algebraically closed field $k$, and let $f$ be a symmetric endomorphism of $A$ (that is, $f^\dagger=f$ where $\dagger$ denotes the Rosati involution). Now, if $p\neq\mbox{char}(k)$ is a prime, $f$ has a representation as an endomorphism of $(T_p A)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$, where $T_pA:=\lim_{\leftarrow}A[p^l]$ and $\mathbb{Z}_p$ (resp. $\mathbb{Q}_p$) denotes the $p$-adic integers (resp. $p$-adic numbers). Moreover, the characteristic polynomial of $f$ when seen as an endomorphism here is characterized by $P_f(t)=\deg(f-t)$ for $t\in\mathbb{Z}$ (see Milne's notes).

My question is the following: If the eigenvalues of (the representation of) $f$ are all integers, does it follow that the minimal polynomial of $f$ splits into different linear factors?

Observation: Over the complex numbers this is true, since the fact that $f$ is symmetric under $\dagger$ means that the analytical representation of $f$ is self-adjoint with respect to the Hermitian form $H=c_1(\mathcal{O}_A(\Theta))$, and so is diagonalizable.

Observation 2: Note that in the first observation the integer valued eigenvalues were never used. I don't know if this hypothesis needs to be used (I doubt it), but the case I'm interested in requires only integer eigenvalues. It may simplify things.

Let $(A,\Theta)$ be a principally polarized abelian variety over an algebraically closed field $k$, and let $f$ be a symmetric endomorphism of $A$ (that is, $f^\dagger=f$ where $\dagger$ denotes the Rosati involution). Now, if $p\neq\mbox{char}(k)$ is a prime, $f$ has a representation as an endomorphism of $(T_p A)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$, where $T_pA:=\lim_{\leftarrow}A[p^l]$ and $\mathbb{Z}_p$ (resp. $\mathbb{Q}_p$) denotes the $p$-adic integers (resp. $p$-adic numbers). Moreover, the characteristic polynomial of $f$ when seen as an endomorphism here is characterized by $P_f(t)=\deg(f-t)$ for $t\in\mathbb{Z}$ (see Milne's notes).

My question is the following: If the eigenvalues of (the representation of) $f$ are all integers, does it follow that the minimal polynomial of $f$ splits into distinct linear factors (that is, no repeated factors)?

Observation: Over the complex numbers this is true, since the fact that $f$ is symmetric under $\dagger$ means that the analytical representation of $f$ is self-adjoint with respect to the Hermitian form $H=c_1(\mathcal{O}_A(\Theta))$, and so is diagonalizable.

Let $(A,\Theta)$ be a principally polarized abelian variety over an algebraically closed field $k$, and let $f$ be a symmetric endomorphism of $A$ (that is, $f^\dagger=f$ where $\dagger$ denotes the Rosati involution). Now, if $p\neq\mbox{char}(k)$ is a prime, $f$ has a representation as an endomorphism of $(T_p A)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$, where $T_pA:=\lim_{\leftarrow}A[p^l]$ and $\mathbb{Z}_p$ (resp. $\mathbb{Q}_p$) denotes the $p$-adic integers (resp. $p$-adic numbers). Moreover, the characteristic polynomial of $f$ when seen as an endomorphism here is characterized by $P_f(t)=\deg(f-t)$ for $t\in\mathbb{Z}$ (see Milne's notes).

My question is the following: If the eigenvalues of (the representation of) $f$ are all integers, does it follow that the minimal polynomial of $f$ splits into different linear factors?

Observation: Over the complex numbers this is true, since the fact that $f$ is symmetric under $\dagger$ means that the analytical representation of $f$ is self-adjoint with respect to the Hermitian form $H=c_1(\mathcal{O}_A(\Theta))$, and so is diagonalizable.

Let $(A,\Theta)$ be a principally polarized abelian variety over an algebraically closed field $k$, and let $f$ be a symmetric endomorphism of $A$ (that is, $f^\dagger=f$ where $\dagger$ denotes the Rosati involution). Now, if $p\neq\mbox{char}(k)$ is a prime, $f$ has a representation as an endomorphism of $(T_p A)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$, where $T_pA:=\lim_{\leftarrow}A[p^l]$ and $\mathbb{Z}_p$ (resp. $\mathbb{Q}_p$) denotes the $p$-adic integers (resp. $p$-adic numbers). Moreover, the characteristic polynomial of $f$ when seen as an endomorphism here is characterized by $P_f(t)=\deg(f-t)$ for $t\in\mathbb{Z}$ (see Milne's notes).

My question is the following: If the eigenvalues of (the representation of) $f$ are all integers, does it follow that the minimal polynomial of $f$ splits into different linear factors?

Observation: Over the complex numbers this is true, since the fact that $f$ is symmetric under $\dagger$ means that the analytical representation of $f$ is self-adjoint with respect to the Hermitian form $H=c_1(\mathcal{O}_A(\Theta))$, and so is diagonalizable.

Observation 2: Note that in the first observation the integer valued eigenvalues were never used. I don't know if this hypothesis needs to be used (I doubt it), but the case I'm interested in requires only integer eigenvalues. It may simplify things.