Find the monic generator of and ideal. Let $\mathbb{F}$ be a subfield of complex numbers, and let
$$
A = \begin{bmatrix}
       1  & -2  \\[0.1em]
       0  &  3  \\[0.1em]
   \end{bmatrix}
$$
Find the monic generator of the ideal of all polynomials $f$ in $F[x]$ such that $f(A)=0.$
 A: That polynomial is called the minimal polynomial. Notice that $1$ and $3$ are the eigenvalues. They are $2$ and they are different. Therefore the characteristic polynomial is the minimal polynomial. 
$f(x)=(x-1)(x-3)$.
Alternative not knowing much.
We need $f(x)=x^n+a_{n-1}x^{n-1}+...a_1x+a_0$ such that $f(A)=0$. An equation like $f(A)=0$ will give us four ordinary equations when we equate the corresponding components. Compute a few of the powers $I=A^0$, $A=A^1$, $A^2$, ... and notice that already $I,A,A^2$ are linearly dependent. Solve the system of equations $A^2+a_1A+a_0=0$ for the unknowns $a_1,a_0$.
You will get $a_1=4$ and $a_0=3$.
Notice that $I$ and $A$ are linearly independent. Therefore there is no such polynomial of degree $1$.
A: We can solve this in a reasonably simple manner.
Consider some $f \in F[x]$ such that $f(A) = 0$. Suppose that $deg(f) = 1$. Then, $f$ is of the form $$ f(x) = cx + d. $$ However, we need $f(A)$ to be zero and all values are in $\mathbb R^{2x2}$, so
\begin{align*}
    f(A) &= cA + d \\
    &= c\begin{bmatrix}1&-2\\0&3\end{bmatrix} + \begin{bmatrix}d&0\\0&d\end{bmatrix} \\
    &= \begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}
\end{align*}
So, unless we set $c=0$ and $d=0$, we cannot scale or add by any $c$ and $d$ such that $f(A)=0$. We can, however, try a function of degree 2, written as $$f(x)=x^2+cx+d.$$
Rewriting a bit, we see that
\begin{align*}
    0 &= f(A) \\
    &= A^2 + cA + d \\
    &= \begin{bmatrix}1&-8\\0&9\end{bmatrix}+\begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}\\
    &= \begin{bmatrix}1+c+d&-8-2c\\0&9+3c+d\end{bmatrix}
\end{align*}
Because each entry of $f(A)$ needs to be zero, we can find $c$ and $d$ by solving this system of linear equations:
$$
\begin{cases}
-1=c+d \\-8=2c \\-9=3c+d
\end{cases}
$$
This is solved by $c=-4$ and substituting into either equation to find that $d=3$. Thus, we have that $$ f(x)=x^2-4x+3 $$ and that $$ f(A)=A^2-4A+3 = 0, $$ as desired. Thus, $f$ is the monic polynomial that generates the ideal $fF[x]$; as such, $f$ is the gcd for all polynomials $g$ such that $g(A)=0$.
(Note that I am implicitly applying the isomorphism from the set of polynomials over $F$, denoted as $F[x]$, to the set of all polynomial functions over $F$, which I usually denote as $\mathbb P_\infty$.)
