The exercise I'm doing is:
Let $a, b, c$ be elements of a field $F$ and let $A$ be the following matrix over $F$. Prove that the characteristic polynomial of $A$ is $x^3-ax^2-bx-c$, and that this is also the minimal polynomial.
$$A=\begin{bmatrix}0&0&c\\1&0&b\\0&1&a\end{bmatrix}$$
So I worked it out as follows:
The characteristic polynomial is defined as $f=det (xI-A)$.$$f=det\begin{bmatrix}x&0&-c\\-1&x&-b\\0&-1&x-a\end{bmatrix}=x^3-ax^2-bx-c$$(Too much typing to write everything out.)
Since $a, b, c$ are yet undetermined, this equation has the potential of having 3 unique solutions. Therefore we require the minimal polynomial to be of the third degree and thus, this polynomial is also the minimum polynomial.
Now, this is all technically correct, but I'm not sure if the last argument is rigorous enough. How would one go about making that fully rigorous?
I agree - the answer is correct but it's hard to see a way of making this argument rigorous.
Can I suggest an alternative approach? Let's focus on the vector $$e_1 = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]$$
Suppose, for contradiction, that there exists a non-zero polynomial of degree less than three, $$c_0 + c_1 A + c_2 A^2$$that annihilates every vector in the vector space. Then in particular, it must annihilate $e_1$. But $$ (c_0 + c_1 A + c_2 A^2) e_1 = \left[ \begin{array}{c} c_0 \\ c_1 \\ c_2 \end{array} \right].$$ so $c_0 + c_1 A + c_2 A^2$ can't annihilate $e_1$ (unless $c_0 = c_1 = c_2 = 0$). Hence the minimal polynomial must be at least a cubic!
Finally, I would invite you to think about how this idea might generalise for arbitrary matrices written in rational canonical form.