Show that the characteristic polynomial is the same as the minimal polynomial

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

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.

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 $1$.