Characteristic polynomial equals minimal polynomial iff $x, Mx, \ldots, M^{n-1} x$ are linearly independent I have been trying to compile conditions for when characteristic polynomials equal minimal polynomials and I have found a result that I think is fairly standard but I have not been able to come up with a proof for.  Any references to a proof would be greatly appreciated.
Let $M\in M_n(\Bbb F)$ and let $c_M$ be the characteristic polynomial of $M$ and $p_M$ be the minimal polynomial of $M$.

How do we show that $p_M = c_M$ if and only if there exists a column vector $x$ such that $x, Mx, \ldots M^{n-1} x$ are linearly independent ?


Here is a simple proof for the "only if" part, using rational canonical form.
For clarity's sake, I'll assume that $M$ is a linear map.
If $M$ is such that $p_M=c_M$, then it is similar to $F$, the companion matrix of $p_M$.i.e. There is a basis $\beta = (v_1, \dots ,v_n)$ for $V$ under which the matrix of $M$ is $F$. Let $e_i (i \in \{1,\dots,n\})$ be the vector with $1$ as its $i$-th component and $0$ anywhere else, then $F^{i} e_{1} = e_{i+1}$ for $i \in \{1, \dots , n-1\}$. But $e_1, \dots , e_n$ are just the coordinates of the vectors $v_1, \dots , v_n$ under the basis $\beta$ itself. So this means that $M^i v_1 = v_{i+1}$ for $i \in \{1, \dots , n-1\}$. In other words, $(v_1, M v_1, \dots , M^{n-1} v_1) = (v_1, v_2, \dots , v_n)$ is a basis for $V$, which is of course linearly independent.
