characteristic polynomial of semisimple matrix Is the following true:

If $A,B$ are two semisimple square matrices over a finite field
  with equal characteristic polynomials, then $A,B$ are similar, that is, there exists an invertible matrix $T$ such that $T^{-1}AT=B$.

 A: Let $A$ be a semisimple endomorphism of an $n$-dimensional vector space $V$ over some field $F$. Then $V$ can be written as direct sum of simple invariant subspaces, corresponding to irreducible factors of the characteristic polynomial $\chi_A(X)=X^n+a_{n-1}X^{n-1}+\ldots +a_0\in F[X]$ of $A$. Since we are interested in $A,B$ having the same characteritic polynomial, the irreducible factors are the same for both and it suffices to consider the corresponding simple subspaces, i.e., we may assume that $\chi_A$ is irreducible. Then $\chi_A$ is also the minimal polynomial.
Pick $0\ne v_0\in V$ and for $i=1,\ldots,n-1$ let $v_{i}=Av_{i-1}$. As $\langle v_0,\ldots,v_{n-1}\rangle$ is $A$-invariant and nonzero, we conclude that $v_0,\ldots, v_{n-1}$ is a basis of $V$. With respect to this basis, $A$ becomes
$$\begin{pmatrix}0&0&\cdots&0&-a_0\\
1&0&\cdots &0&-a_1\\
0&1&\cdots&0&-a_2\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&\cdots&1&-a_{n-1}\end{pmatrix} $$
Since the same argument works for $B$, we conclude that $A\sim B$.
