Let $T$ be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that $T$ is diagonalizable if and only if $T$ is annihilated by some polynomial over $\Bbb{C}$ which has distinct roots.
This problem is bit vague for me. Is “polynomial over $\Bbb{C}$ which has distinct roots” means polynomial of form $(x-c_1)\cdots (x-c_k)$, where $c_1,…,c_k\in \Bbb{C}$ are distinct?
Proof: $(\Rightarrow)$ Suppose $T$ is diagonalizable. Let $m$ be minimal polynomial of $T$. By theorem 6 section 6.4, $m=(x-c_1)\cdots (x-c_k)$, where $c_1,…,c_k\in \Bbb{C}$ are distinct. $(\Leftarrow)$ Conversely, suppose $\exists f\in F[x]$ such that $f=(x-c_1)\cdots (x-c_k)$, where $c_1,…,c_k\in \Bbb{C}$ are distinct and $f(T)=0$. Then $m|f$. So $m= (x-c_1)^{e_1}\cdots (x-c_k)^{e_k}$; $0\leq e_i\leq 1$. By theorem 6 section 6.4, $T$ is diagonalizable.
In proof, we didn’t use any property of $\Bbb{C}$. So this problem holds in field $F$. Am I right?