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Let $$A =beginpmatrix0 & 0 & c \1 & 0 & b \ 0& 1 & aendpmatrix$$
I wish to show that the characteristic and minimal polynomials are the same.

I have already found via computation that the characteristic polynomial is $p_A(x)=x^3-ax^2-bx-c$, and I know from here that if I could show that the eigenspaces of $A$ all have dimension $1$, I'd be done.

The problem is that, solving for the eigenvalues of this (very general) cubic equation is difficult (though, possible), meaning it would be difficult to find bases for the eigenspaces.

A hint would be appreciated.

Compute:
$$A^2 = beginpmatrix 0 & c & ac \ 0 & b & c + ab \ 1 & a & b + a^2endpmatrix.$$
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
$$A^2 = pA + qI iff beginpmatrix 0 & c & ac \ 0 & b & c + ab \ 1 & a & b + a^2endpmatrix = pbeginpmatrix 0 & 0 & c \ 1 & 0 & b \ 0 & 1 & cendpmatrix + qbeginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1endpmatrix$$
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 $ntimes n$ matrix of the polynomial $$p(x)=x^n-c_n-1x^n-1-cdots-c_1x-c_0.$$ 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 $p_A(A) = O_3times 3$ 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 $m(A) = O_3times 3$.

Applying $m(A)$ to $e_1$ gives
$$m(A)e_1 = A^2e_1 + uAe_1 + ve_1 = e_3 +ue_2 + ve_1 = beginpmatrix0 \ 0 \ 0endpmatrix mbox Contradiction!$$
The linear combination cannot result in the zero vector as the coefficient of the basis vector $e_3$ is $1$.