Show that the characteristic polynomial is the same as the minimal polynomial 
Let $$A =\begin{pmatrix}0 & 0 & c \\1 & 0 & b \\ 0& 1 & a\end{pmatrix}$$
  Show that the characteristic and minimal polynomials of $A$ are the same. 

I have already computated the characteristic polynomial 
$$p_A(x)=x^3-ax^2-bx-c$$
and I know from here that if I could show that the eigenspaces of $A$ all have dimension $1$, I would be done. The problem is that solving for the eigenvalues of this (very general) cubic equation is difficult (albeit possible), meaning it would be difficult to find bases for the eigenspaces.
A hint would be appreciated. 
 A: Compute:
$$A^2 = \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix}.$$
So, we just need to show that $A^2, A, I$ are linearly independent. Clearly $A$ is not a multiple of $I$, so we just need to show there is no solution to the equation
$$A^2 = pA + qI \iff \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix} = p\begin{pmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a\end{pmatrix} + q\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
for $p$ and $q$. In particular, if you examine the entries in the left column, bottom row, we get
$$1 = 0p + 0q,$$
which means there is indeed no solution. Hence $I, A, A^2$ are linearly independent, so no quadratic of $A$ will be equal to the $0$ matrix. Thus, the minimal polynomial must be (at least) a cubic, and equal to the characteristic polynomial
A: The form of $A$ has a special name: the companion matrix of the polynomial $p(x)=x^3-ax^2-bx-c$.  
For the standard basis $e_1,e_2,e_3$, one finds that $Ae_1=e_2$, $Ae_2=e_3$, so $\{e_1,Ae_1,A^2e_1\}$ forms a basis. 
The general context is the companion $n\times n$ matrix of the polynomial $$p(x)=x^n-c_{n-1}x^{n-1}-\cdots-c_1x-c_0.$$ A  vector $v$ is said to be a cyclic vector for $A$ if the iterates by $A$ of $v$ for a basis for $R^n$. As others point out, this suffices to show that the minimal polynomial is the same as the characteristic polynomial.
A: Assuming you know already that according to Cayley-Hamilton you have $p_A(A) = O_{3\times 3}$ you can also proceed as follows:


*

*Let $e_1, e_2, e_3$ denote the canonical basis $\Rightarrow Ae_1=e_2, Ae_2 = e_3 \Rightarrow A^2e_1 = e_3$
Now, assume there is a polynomial $m(x)=x^2+ux+v$ such that $m(A) = O_{3\times 3}$.
Applying $m(A)$ to $e_1$ gives
$$m(A)e_1 = A^2e_1 + uAe_1 + ve_1 = e_3 +ue_2 + ve_1 = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \mbox{ Contradiction!}$$
The linear combination cannot result in the zero vector as the coefficient of the basis vector $e_3$ is $1$.
